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Thermometry, Thermal Expansion and Calorimetry

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130
MHT CET
42
JEE Main
79
NEET
9
Years
1988–2026
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Q1MHT CET 2026 · 19 MAY · SHIFT 1Easy

Thermometry & Thermal Expansion · Temperature scales & thermometers

The length of mercury column in a mercury-in-glass thermometer is H1H_1 mm at the ice point and H2H_2 mm at the steam point. When the length is HH mm, the temperature TT is given in ∘^\circC by (H1<H2H_1 < H_2)

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Answer: (a)

  1. Mercury length rises linearly with temperature: T−0100−0=H−H1H2−H1\frac{T-0}{100-0}=\frac{H-H_1}{H_2-H_1}.
  2. So the answer is (a) T=100(H−H1)H2−H1T=\frac{100(H-H_1)}{H_2-H_1}.
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Q2MHT CET 2026 · 19 APR · SHIFT 1Easy

Thermometry & Thermal Expansion · Temperature scales & thermometers

If the two temperatures on Fahrenheit scale differ by 54∘54^\circF, then the difference in temperature on the Celsius scale is

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Answer: (a)

  1. 100100 C divisions =180=180 F divisions, so ΔC=59ΔF\Delta C=\frac{5}{9}\Delta F.
  2. ΔC=59×54=30\Delta C=\frac{5}{9}\times 54=30
  3. So the answer is (a) 30∘30^\circC.
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Q3MHT CET 2026 · 12 MAY · SHIFT 2Easy

Thermometry & Thermal Expansion · Area & volume expansion

What is the change in volume of an iron sphere of volume 400400 cm3^3, when it is heated from 0∘C0^\circ\text{C} to 50∘C50^\circ\text{C}? (αiron=12×10−6/∘C\alpha_{\text{iron}} = 12 \times 10^{-6}/^\circ\text{C})

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Answer: (c)

  1. γ=3α=36×10−6/∘\gamma=3\alpha=36\times10^{-6}/^\circC, ΔV=VγΔT\Delta V=V\gamma\Delta T
  2. ΔV=400×36×10−6×50=0.72\Delta V=400\times 36\times10^{-6}\times 50=0.72 cm3^3
  3. So the answer is (c) 0.720.72 cm3^3.
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Q4NEET 2026 · RE-NEETEasy

Thermometry & Thermal Expansion · Area & volume expansion

The temperature of a metallic sphere of radius RR is increased by a small amount ΔT\Delta T. If the linear coefficient of thermal expansion of the metal is α\alpha, the approximate increase in the volume of the sphere is:

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Answer: (d)

  1. V=43πR3V=\frac{4}{3}\pi R^3 and γ=3α\gamma=3\alpha.
  2. ΔV=VγΔT=43πR3×3αΔT=4πR3αΔT\Delta V=V\gamma\Delta T=\frac{4}{3}\pi R^3\times 3\alpha\Delta T=4\pi R^3\alpha\Delta T
  3. So the answer is (d).
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Q5MHT CET 2026 · 14 MAY · SHIFT 1Medium

Thermometry & Thermal Expansion · Temperature scales & thermometers

The coefficient of volume expansion of mercury is 18×10−5/∘18 \times 10^{-5}/^\circC. A thermometer bulb has volume 10−610^{-6} m3^3 and cross-section of stem is 0.0040.004 cm2^2. Assuming bulb is filled with mercury at 0∘0^\circC then the mercury column at 100∘100^\circC is

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Answer: (d)

  1. Rise in mercury volume (bulb expansion neglected): ΔV=VγΔT=10−6×18×10−5×100=1.8×10−8\Delta V=V\gamma\Delta T=10^{-6}\times18\times10^{-5}\times100=1.8\times10^{-8} m3^3
  2. Stem area A=0.004A=0.004 cm2=4×10−7^2=4\times10^{-7} m2^2
  3. l=ΔVA=1.8×10−84×10−7=0.045l=\frac{\Delta V}{A}=\frac{1.8\times10^{-8}}{4\times10^{-7}}=0.045 m
  4. So the answer is (d) 4.54.5 cm.
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Q6MHT CET 2026 · 13 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Temperature scales & thermometers

A celsius and a Fahrenheit thermometer are dipped in boiling water. The temperature of water is lowered until the Farenheit thermometer shows 140∘140^\circ. What is the fall in temperature on the Celsius thermometer?

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Answer: (b)

  1. Boiling water: 100∘100^\circC. At 140∘140^\circF: C=59(140−32)=60∘C=\frac{5}{9}(140-32)=60^\circC.
  2. Fall =100−60=40∘=100-60=40^\circC
  3. So the answer is (b) 40∘40^\circC.
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Q7MHT CET 2026 · 15 MAY · SHIFT 1Medium

Thermometry & Thermal Expansion · Linear expansion

An iron tyre is to be fitted on to a wooden wheel 1.01.0 m in diameter. The diameter of the tyre is 66 mm smaller than that of the wheel. The tyre should be heated so that its temperature increases by a minimum of

(coefficient of volume expansion of iron = 3.6×10−5/∘3.6 \times 10^{-5}/^\circC)

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Answer: (c)

  1. α=γ3=1.2×10−5/∘\alpha=\frac{\gamma}{3}=1.2\times10^{-5}/^\circC; the diameter must grow by 66 mm.
  2. ΔT=Δdd α=6×10−31.0×1.2×10−5=500∘\Delta T=\frac{\Delta d}{d\,\alpha}=\frac{6\times10^{-3}}{1.0\times1.2\times10^{-5}}=500^\circC
  3. So the answer is (c) 500∘500^\circC.
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Q8JEE Main 2026 · 4 APR · SHIFT 2Medium

Thermometry & Thermal Expansion · Linear expansion

The temperature of a metal strip having coefficient of linear expansion α\alpha is increased from T1T_1 to T2T_2 resulting in increase of its length by ΔL1\Delta L_1. The temperature is further increased from T2T_2 to T3T_3 such that the increase in its length is ΔL2\Delta L_2.

Given T3+T1=2T2T_3 + T_1 = 2T_2 and T2−T1=ΔTT_2 - T_1 = \Delta T, the value of ΔL2\Delta L_2 is ______.

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Answer: (d)

  1. ΔL1=L1αΔT\Delta L_1 = L_1\alpha\Delta T and the length at T2T_2 is L2=L1(1+αΔT)L_2 = L_1(1+\alpha\Delta T).
  2. From T3+T1=2T2T_3 + T_1 = 2T_2: T3−T2=T2−T1=ΔTT_3 - T_2 = T_2 - T_1 = \Delta T.
  3. ΔL2=L2αΔT=L1αΔT(1+αΔT)\Delta L_2 = L_2\alpha\Delta T = L_1\alpha\Delta T(1+\alpha\Delta T)
  4. So the answer is (d) ΔL1[1+αΔT]\Delta L_1[1+\alpha\Delta T].
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Q9JEE Main 2026 · 21 JAN · SHIFT 1Medium

Thermometry & Thermal Expansion · Linear expansion

An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at 30∘C30^\circ\text{C}. The coefficient of linear expansion of aluminium and steel are 24×10−6/∘C24 \times 10^{-6}/^\circ\text{C} and 1.2×10−5/∘C1.2 \times 10^{-5}/^\circ\text{C}, respectively. The length of this composite rod when its temperature is raised to 100∘C100^\circ\text{C}, is ______ cm.

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Answer: (c)

  1. Each rod is 6060 cm long; Δθ=100−30=70∘C\Delta\theta = 100 - 30 = 70^\circ\text{C}.
  2. ΔL=60×70×(24+12)×10−6=0.1512\Delta L = 60 \times 70 \times (24 + 12)\times 10^{-6} = 0.1512 cm
  3. L=120+0.15=120.15L = 120 + 0.15 = 120.15 cm
  4. So the answer is (c) 120.15 cm.
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Q10MHT CET 2026 · 15 MAY · SHIFT 2Medium

Thermometry & Thermal Expansion · Area & volume expansion

A sheet of solid material having area AA and coefficient of linear expansion α\alpha. The mass of sheet is mm and specific heat capacity is ss. If dQdQ is the amount of heat given to the sheet then change in its area is

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Answer: (b)

  1. Heat raises temperature by ΔT=ΔQms\Delta T = \dfrac{\Delta Q}{ms}.
  2. Areal expansion: ΔA=A(2α)ΔT\Delta A = A(2\alpha)\Delta T.
  3. ΔA=A(2α)ΔQms\Delta A = \dfrac{A(2\alpha)\Delta Q}{ms}, so the answer is (b).
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Q11JEE Main 2026 · 5 APR · SHIFT 2Medium

Thermometry & Thermal Expansion · Area & volume expansion

An ideal gas at pressure PP and temperature TT is expanding such that PT3=PT^3 = constant. The coefficient of volume expansion of the gas is ______.

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Answer: (c)

  1. PT3=PT^3 = const ⇒P∝T−3\Rightarrow P \propto T^{-3}.
  2. Ideal gas: V∝TP∝T4V \propto \dfrac{T}{P} \propto T^4.
  3. γ=1VdVdT=4T\gamma = \dfrac{1}{V}\dfrac{dV}{dT} = \dfrac{4}{T}, answer (c).
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Q12MHT CET 2026 · 14 MAY · SHIFT 2Difficult

Thermometry & Thermal Expansion · Linear expansion

A metal rod (A) of length 2020 cm expands by 0.0750.075 cm when its temperature changes from 0∘0^{\circ}C to 100∘100^{\circ} C. Another metal rod (B) of different metal of same length expands by 0.0450.045 cm for same temperature change. A third rod of same length is composed of 2 parts, one of metal A and another of metal B. For same temperature change, this rod expands by 0.0600.060 cm. The portion made of metal A has the length

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Answer: (b)

  1. Expansion per cm for 100∘100^\circC: A gives 0.075/20=0.003750.075/20 = 0.00375, B gives 0.045/20=0.002250.045/20 = 0.00225.
  2. Let A part be xx: 0.00375x+0.00225(20−x)=0.0600.00375x + 0.00225(20 - x) = 0.060
  3. 0.0015x=0.015⇒x=100.0015x = 0.015 \Rightarrow x = 10 cm
  4. So the answer is (b) 10 cm.
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Q13MHT CET 2025 · 22 APR · SHIFT 1Easy

Thermometry & Thermal Expansion · Temperature scales & thermometers

The temperature of a body on Kelvin scale is ' xx ' KK. When it is measured by a Fahrenheit thermometer, it is found to be ' x ' ∘F^\circ\text{F}. The value of ' xx ' is (nearly)

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Answer: (c)

  1. K=C+273K=C+273 and C5=F−329\frac{C}{5}=\frac{F-32}{9}, so F=95(K−273)+32F=\frac{9}{5}(K-273)+32.
  2. Put F=K=xF=K=x: 5x=9x−2457+160⇒4x=22975x=9x-2457+160 \Rightarrow 4x=2297
  3. x≈574x\approx 574
  4. So the answer is (c) 574.
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Q14MHT CET 2025 · 21 APR · SHIFT 1Easy

Thermometry & Thermal Expansion · Temperature scales & thermometers

A centigrade and Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit temperature observed is 140∘F140^\circ\text{F}. At that time the temperature registered by the centigrade thermometer is

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Answer: (b)

  1. C5=F−329\frac{C}{5}=\frac{F-32}{9}
  2. C=59(140−32)=59×108=60C=\frac{5}{9}(140-32)=\frac{5}{9}\times 108=60
  3. So the answer is (b) 60∘60^\circC.
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Q15JEE Main 2025 · 24 JAN · SHIFT 2Easy

Thermometry & Thermal Expansion · Temperature scales & thermometers

Which of the following figure represents the relation between Celsius and Fahrenheit temperatures?

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Answer: (d)

  1. C=59(F−32)=59F−1609C=\frac{5}{9}(F-32)=\frac{5}{9}F-\frac{160}{9}: a straight line with positive slope.
  2. At F=0F=0, C=−17.8C=-17.8 (negative intercept); C=0C=0 at F=32F=32 (positive F-intercept).
  3. Only graph (d) rises to the right and cuts the C-axis below the origin.
  4. So the answer is (d).
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Q16MHT CET 2025 · 25 APR · SHIFT 2Easy

Thermometry & Thermal Expansion · Area & volume expansion

The volume of a metal sphere increases by 0.33%0.33\% when its temperature is raised by 50∘C50^\circ\text{C}. The coefficient of linear expansion of the metal is

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Answer: (a)

  1. ΔVV=γΔT=3αΔT\frac{\Delta V}{V}=\gamma\Delta T=3\alpha\Delta T
  2. 0.0033=3α×50⇒α=0.0033150=2.2×10−5/∘0.0033=3\alpha\times 50\Rightarrow \alpha=\frac{0.0033}{150}=2.2\times10^{-5}/^\circC
  3. So the answer is (a).
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Q17MHT CET 2025 · 23 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Temperature scales & thermometers

A thermometer bulb has volume 10−6 m310^{-6}\text{ m}^3 and cross-section of the stem is 0.002 cm20.002\text{ cm}^2. The bulb is filled with mercury at 0∘C0^\circ\text{C}. If the thermometer reads temperature as 100∘C100^\circ\text{C}, then the length of mercury column is (coefficient of cubical expansion of mercury =18×10−5/∘C= 18 \times 10^{-5}/^\circ\text{C})

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Answer: (c)

  1. ΔV=VγΔT=10−6×18×10−5×100=1.8×10−8\Delta V=V\gamma\Delta T=10^{-6}\times18\times10^{-5}\times100=1.8\times10^{-8} m3^3
  2. A=0.002A=0.002 cm2=2×10−7^2=2\times10^{-7} m2^2
  3. l=ΔVA=1.8×10−82×10−7=0.09l=\frac{\Delta V}{A}=\frac{1.8\times10^{-8}}{2\times10^{-7}}=0.09 m
  4. So the answer is (c) 9 cm.
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Q18MHT CET 2025 · 5 MAY · SHIFT 2Medium

Thermometry & Thermal Expansion · Linear expansion

The length of steel rod is 5 cm longer than the copper rod at all temperatures. The length of the steel and copper rod is respectively (Coefficient of linear expansion for steel and copper is respectively 1.1×10−5/∘C1.1 \times 10^{-5}/^\circ\text{C} and 1.7×10−5/∘C1.7 \times 10^{-5}/^\circ\text{C})

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Answer: (b)

  1. Constant difference needs equal expansions: Lsαs=LcαcL_s\alpha_s=L_c\alpha_c.
  2. 1.1Ls=1.7(Ls−5)⇒0.6Ls=8.5⇒Ls≈14.21.1L_s=1.7(L_s-5)\Rightarrow 0.6L_s=8.5\Rightarrow L_s\approx14.2 cm
  3. Lc=Ls−5≈9.2L_c=L_s-5\approx9.2 cm
  4. So the answer is (b) nearly 14 cm and 9 cm.
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Q19MHT CET 2025 · 26 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Linear expansion

The difference in length between two rods A and B is 60 cm at all temperatures. If αA=18×10−6/∘C\alpha_{\text{A}} = 18 \times 10^{-6}/^\circ\text{C} and αB=27×10−6/∘C\alpha_{\text{B}} = 27 \times 10^{-6}/^\circ\text{C}, then the length of rod A\text{rod A} and rod B\text{rod B} at 0∘C0^\circ\text{C} is respectively

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Answer: (b)

  1. Constant difference: lAαA=lBαB⇒18 lA=27 lB⇒lA=1.5 lBl_A\alpha_A=l_B\alpha_B\Rightarrow 18\,l_A=27\,l_B\Rightarrow l_A=1.5\,l_B
  2. lA−lB=0.5 lB=60⇒lB=120l_A-l_B=0.5\,l_B=60\Rightarrow l_B=120 cm, lA=180l_A=180 cm
  3. So the answer is (b).
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Q20MHT CET 2025 · 19 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Linear expansion

Two rods of different materials have lengths ' l1l_1 ' and ' l2l_2 ' whose coefficient of linear expansions are ' α1\alpha_1 ' and ' α2\alpha_2 ' respectively. If the difference between the two lengths is independent of temperature then

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Answer: (b)

  1. Difference constant means equal expansions: l1α1ΔT=l2α2ΔTl_1\alpha_1\Delta T=l_2\alpha_2\Delta T.
  2. l1l2=α2α1\frac{l_1}{l_2}=\frac{\alpha_2}{\alpha_1}
  3. So the answer is (b).
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Q21JEE Main 2025 · 4 APR · SHIFT 2Medium

Thermometry & Thermal Expansion · Area & volume expansion

Consider a rectangular sheet of solid material of length l=9 cml = 9\text{ cm} and width d=4 cm\text{d} = 4\text{ cm}. The coefficient of linear expansion is α=3.1×10−5 K−1\alpha = 3.1 \times 10^{-5}\text{ K}^{-1} at room temperature and one atmospheric pressure. The mass of sheet m=0.1 kgm = 0.1\text{ kg} and the specific heat capacity Cv=900 J kg−1 K−1C_{\text{v}} = 900\text{ J kg}^{-1}\text{ K}^{-1}. If the amount of heat supplied to the material is 8.1×102 J8.1 \times 10^2\text{ J} then change in area of the rectangular sheet is:

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Answer: (a)

  1. ΔT=Qmc=8100.1×900=9\Delta T = \dfrac{Q}{mc} = \dfrac{810}{0.1 \times 900} = 9 K
  2. A=9×4=36 cm2=36×10−4 m2A = 9 \times 4 = 36\text{ cm}^2 = 36\times10^{-4}\text{ m}^2
  3. ΔA=A(2α)ΔT=36×10−4×6.2×10−5×9≈2.0×10−6 m2\Delta A = A(2\alpha)\Delta T = 36\times10^{-4} \times 6.2\times10^{-5} \times 9 \approx 2.0\times10^{-6}\text{ m}^2
  4. So the answer is (a).
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Q22JEE Main 2024 · 4 APR · SHIFT 1Easy

Thermometry & Thermal Expansion · Temperature scales & thermometers

On Celsius scale the temperature of body increases by 40∘C40^\circ\text{C}. The increase in temperature on Fahrenheit scale is:

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Answer: (c)

  1. ΔF=95ΔC\Delta F=\frac{9}{5}\Delta C
  2. ΔF=95×40=72\Delta F=\frac{9}{5}\times 40=72
  3. So the answer is (c) 72∘72^\circF.
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Q23MHT CET 2024 · 11 MAY · SHIFT 2Easy

Thermometry & Thermal Expansion · Linear expansion

A metal rod having coefficient of linear expansion 2×10−5/∘C2 \times 10^{-5}/^\circ\text{C} is 0.75 m long at 45∘C45^\circ\text{C}. When the temperature rises to 65∘C65^\circ\text{C}, the increase in length of the rod will be

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Answer: (c)

  1. ΔL=LαΔT\Delta L=L\alpha\Delta T
  2. ΔL=0.75×2×10−5×20=3×10−4\Delta L=0.75\times 2\times10^{-5}\times 20=3\times10^{-4} m
  3. So the answer is (c) 0.30 mm.
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Q24JEE Main 2024 · 4 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Temperature scales & thermometers

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are 8 Ω8\ \Omega and 10 Ω10\ \Omega respectively. After inserting in a hot bath of temperature 400∘C400^\circ\text{C}, the resistance of platinum wire is:

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Answer: (b)

  1. t100=Rt−R0R100−R0\frac{t}{100}=\frac{R_t-R_0}{R_{100}-R_0}
  2. 400100=R−810−8⇒R−8=8\frac{400}{100}=\frac{R-8}{10-8}\Rightarrow R-8=8
  3. R=16 ΩR=16\ \Omega
  4. So the answer is (b) 16 Ω\Omega.
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Q25MHT CET 2024 · 16 MAY · SHIFT 2Medium

Thermometry & Thermal Expansion · Linear expansion

At certain temperature, rod A\text{rod A} and rod B\text{rod B} of different materials have lengths LA\text{L}_{\text{A}} and LB\text{L}_B respectively. Their co-efficients of linear expansion are αA\alpha_A and αB\alpha_B respectively. It is observed that the difference between their lengths remain constant at all temperatures. The ratio LA/LBL_A/L_B is given by

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Answer: (b)

  1. Difference constant means LAαAΔT=LBαBΔTL_A\alpha_A\Delta T=L_B\alpha_B\Delta T.
  2. LALB=αBαA\frac{L_A}{L_B}=\frac{\alpha_B}{\alpha_A}
  3. So the answer is (b).
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Q26MHT CET 2024 · 11 MAY · SHIFT 1Medium

Thermometry & Thermal Expansion · Linear expansion

Rails of material of steel are laid with gaps to allow for thermal expansion. Each track is 10 m long, when laid at temperature 17∘C17^\circ\text{C}. The maximum temperature that can be reached is 45∘C45^\circ\text{C}. The gap to be kept between the two segments of railway track is

(α(\alphasteel =1.3×10−5/∘C)= 1.3 \times 10^{-5}/^\circ\text{C})

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Answer: (c)

  1. Gap == expansion of one segment: ΔL=LαΔT\Delta L=L\alpha\Delta T
  2. ΔL=10×1.3×10−5×(45−17)=3.64×10−3\Delta L=10\times1.3\times10^{-5}\times(45-17)=3.64\times10^{-3} m
  3. So the answer is (c) 3.64 mm.
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Q27MHT CET 2024 · 9 MAY · SHIFT 1Medium

Thermometry & Thermal Expansion · Linear expansion

At certain temperature, rod A\text{rod A} and rod B\text{rod B} of different materials have lengths LA\text{L}_{\text{A}} and LB\text{L}_{\text{B}} respectively. Their coefficients of linear expansion are αA\alpha_A and αB\alpha_B respectively. It is observed that the difference between their lengths remains constant at all temperatures. The ratio LA:LB\text{L}_{\text{A}} : \text{L}_{\text{B}} is given by

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Answer: (b)

  1. Difference constant means LAαAΔT=LBαBΔTL_A\alpha_A\Delta T=L_B\alpha_B\Delta T.
  2. LALB=αBαA\frac{L_A}{L_B}=\frac{\alpha_B}{\alpha_A}
  3. So the answer is (b).
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Q28MHT CET 2024 · 2 MAY · SHIFT 1Medium

Thermometry & Thermal Expansion · Linear expansion

Railway track is made of steel segments separated by small gaps to allow for linear expansion. The segment of track is 10 m long when laid at temperature 17∘C17^\circ\text{C}. The maximum temperature that can be reached is 45∘C45^\circ\text{C}. Increase in length of the segment of railway track is ' xx ' ×10−5 m\times 10^{-5}\text{ m}. The value of ' xx ' is (α(\alphasteel =1.2×10−5/∘C)= 1.2 \times 10^{-5}/^\circ\text{C})

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Answer: (c)

  1. ΔL=LαΔT=10×1.2×10−5×28\Delta L=L\alpha\Delta T=10\times1.2\times10^{-5}\times28
  2. ΔL=336×10−5\Delta L=336\times10^{-5} m
  3. So x=336x=336, answer (c).
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Q29MHT CET 2024 · 15 MAY · SHIFT 1Medium

Thermometry & Thermal Expansion · Area & volume expansion

A sheet of steel is 40 cm long and 5 cm broad at 0∘C0^\circ\text{C}. The surface area of the sheet increases by 1.4 cm21.4\text{ cm}^2 at 100∘C100^\circ\text{C}. Coefficient of linear expansion of steel is

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Answer: (c)

  1. A=40×5=200 cm2A = 40 \times 5 = 200\text{ cm}^2, β=2α\beta = 2\alpha.
  2. ΔA=A(2α)ΔT⇒1.4=200×2α×100\Delta A = A(2\alpha)\Delta T \Rightarrow 1.4 = 200 \times 2\alpha \times 100
  3. α=1.440000=3.5×10−5/∘C\alpha = \dfrac{1.4}{40000} = 3.5\times10^{-5}/^\circ\text{C}, answer (c).
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Q30MHT CET 2024 · 3 MAY · SHIFT 2Difficult

Thermometry & Thermal Expansion · Linear expansion

Two rods, one of copper ( Cu)) and the other of iron ( Fe) having initial lengths L1L_1 and L2L_2 respectively are connected together to form a single rod of length L1+L2L_1 + L_2. The coefficient of linear expansion of Cu and Fe are αc\alpha_c and αi\alpha_i respectively. If the length of each rod increases by the same amount when their temperatures are raised by t∘Ct^{\circ}\text{C}, then ratio of L1−L2L1+L2\frac{L_1 - L_2}{L_1 + L_2} will be

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Answer: (c)

  1. Equal expansion: L1αc=L2αi⇒L1L2=αiαcL_1\alpha_c = L_2\alpha_i \Rightarrow \dfrac{L_1}{L_2} = \dfrac{\alpha_i}{\alpha_c}.
  2. L1−L2L1+L2=αi−αcαi+αc\dfrac{L_1 - L_2}{L_1 + L_2} = \dfrac{\alpha_i - \alpha_c}{\alpha_i + \alpha_c}
  3. So the answer is (c).
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Q31MHT CET 2024 · 3 MAY · SHIFT 1Difficult

Thermometry & Thermal Expansion · Linear expansion

Two rods, one of aluminium and the other of steel, having initial lengths ' L1L_1 ' and ' L2L_2 ' are connected together to form a single rod of length (L1+L2)(L_1 + L_2). The coefficients of linear expansion of aluminium and steel are ' α1\alpha_1 ' and ' α2\alpha_2 ' respectively. If the length of each rod increases by the same amount, when their temperatures are raised by t∘Ct^{\circ}\text{C}, then the ratio L1L1+L2\frac{L_1}{L_1 + L_2} will be

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Answer: (c)

  1. Equal expansion: L1α1=L2α2⇒L1L2=α2α1L_1\alpha_1 = L_2\alpha_2 \Rightarrow \dfrac{L_1}{L_2} = \dfrac{\alpha_2}{\alpha_1}.
  2. L1L1+L2=α2α1+α2\dfrac{L_1}{L_1 + L_2} = \dfrac{\alpha_2}{\alpha_1 + \alpha_2}
  3. So the answer is (c).
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Q32MHT CET 2023 · 10 MAY · SHIFT 2Easy

Thermometry & Thermal Expansion · Linear expansion

A metal rod 2 m2\text{ m} long increases in length by 1.6 mm1.6\text{ mm}, when heated from 0∘C0^\circ\text{C} to 60∘C60^\circ\text{C}. The coefficient of linear expansion of metal rod is

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Answer: (a)

  1. α=ΔLLΔT\alpha=\frac{\Delta L}{L\Delta T}
  2. α=1.6×10−32×60=1.33×10−5/∘\alpha=\frac{1.6\times10^{-3}}{2\times 60}=1.33\times10^{-5}/^\circC
  3. So the answer is (a).
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Q33MHT CET 2023 · 12 MAY · SHIFT 2Easy

Thermometry & Thermal Expansion · Area & volume expansion

The volume of a metal block increases by 0.225%0.225\% when its temperature is increased by 30∘C30^\circ\text{C}. Hence coefficient of linear expansion of the material of metal block is

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Answer: (c)

  1. ΔVV=3αΔT\frac{\Delta V}{V}=3\alpha\Delta T
  2. 0.00225=3α×30⇒α=0.0022590=2.5×10−5/∘0.00225=3\alpha\times 30\Rightarrow \alpha=\frac{0.00225}{90}=2.5\times10^{-5}/^\circC
  3. So the answer is (c).
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Q34JEE Main 2023 · 11 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Temperature scales & thermometers

On a temperature scale 'X\text{X}', the boiling point of water is 65∘X65^\circ\text{X} and the freezing point is −15∘X-15^\circ\text{X}. Assume that the X\text{X} scale is linear. The equivalent temperature corresponding to −95∘X-95^\circ\text{X} on the Fahrenheit scale would be:

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Answer: (a)

  1. X−(−15)65−(−15)=C100\frac{X-(-15)}{65-(-15)}=\frac{C}{100}
  2. −95+1580=C100⇒C=−100∘\frac{-95+15}{80}=\frac{C}{100}\Rightarrow C=-100^\circC
  3. F=95(−100)+32=−148∘F=\frac{9}{5}(-100)+32=-148^\circF
  4. So the answer is (a).
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Q35JEE Main 2023 · 25 JAN · SHIFT 2Medium

Thermometry & Thermal Expansion · Temperature scales & thermometers

The graph between two temperature scales P and Q is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal divisions of scale P and 100 divisions on scale Q. The relationship for conversion between the two scales is given by :-

Figure

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Answer: (d)

  1. From the graph, P runs from 3030 (lower point, since 180−150=30180-150=30) to 180180; Q runs from 00 to 100100.
  2. tQ−0100=tP−30150\frac{t_Q-0}{100}=\frac{t_P-30}{150}
  3. So the answer is (d).
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Q36JEE Main 2023 · 30 JAN · SHIFT 2Medium

Thermometry & Thermal Expansion · Temperature scales & thermometers

A faulty thermometer reads 5∘C5^\circ\text{C} in melting ice and 95∘C95^\circ\text{C} in steam. The correct temperature on absolute scale will be __________ K\text{K} when the faulty thermometer reads 41∘C41^\circ\text{C}.

Numerical answer type. Work it out, then open the answer.

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Answer: 313

  1. C100=x−595−5\frac{C}{100}=\frac{x-5}{95-5}
  2. C=41−590×100=40∘C=\frac{41-5}{90}\times100=40^\circC
  3. T=40+273=313T=40+273=313 K
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Q37MHT CET 2023 · 11 MAY · SHIFT 2Medium

Thermometry & Thermal Expansion · Linear expansion

The difference in length between two rods A\text{A} and B\text{B} is 60 cm60\text{ cm} at all temperatures. If αA=18×10−6/∘C\alpha_{\text{A}} = 18 \times 10^{-6}/^\circ\text{C} and αB=27×10−6/∘C\alpha_{\text{B}} = 27 \times 10^{-6}/^\circ\text{C}, the lengths of the two rods are

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Answer: (b)

  1. Constant difference: lAαA=lBαB⇒18 lA=27 lB⇒lA=1.5 lBl_A\alpha_A=l_B\alpha_B\Rightarrow 18\,l_A=27\,l_B\Rightarrow l_A=1.5\,l_B
  2. lA−lB=0.5 lB=60⇒lB=120l_A-l_B=0.5\,l_B=60\Rightarrow l_B=120 cm, lA=180l_A=180 cm
  3. So the answer is (b).
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Q38MHT CET 2023 · 9 MAY · SHIFT 2Medium

Thermometry & Thermal Expansion · Area & volume expansion

The side of a copper cube is 1 m1\text{ m} at 0∘C0^\circ\text{C}. What will be the change in its volume, when it is heated to 100∘C100^\circ\text{C}? [α[\alphacopper =18×10−6/∘C]= 18 \times 10^{-6}/^\circ\text{C}]

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Answer: (b)

  1. ΔV=V(3α)ΔT\Delta V = V(3\alpha)\Delta T with V=1 m3V = 1\text{ m}^3.
  2. ΔV=1×3×18×10−6×100=5.4×10−3 m3\Delta V = 1 \times 3 \times 18\times10^{-6} \times 100 = 5.4\times10^{-3}\text{ m}^3
  3. So the answer is (b) 54×10−4 m354\times10^{-4}\text{ m}^3.
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Q39MHT CET 2023 · 9 MAY · SHIFT 1Medium

Thermometry & Thermal Expansion · Area & volume expansion

Two uniform brass rods AA and BB of length 'll' and '2l2l' and their radii '2r2r' and 'rr' respectively are heated to same temperature. The ratio of the increase in the volume of rod A\text{rod A} to that of rod B\text{rod B} is

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Answer: (c)

  1. Same material and same ΔT\Delta T, so ΔV∝V\Delta V \propto V.
  2. VA=π(2r)2l=4πr2lV_A = \pi(2r)^2 l = 4\pi r^2 l, VB=πr2(2l)=2πr2lV_B = \pi r^2 (2l) = 2\pi r^2 l
  3. ΔVA:ΔVB=2:1\Delta V_A : \Delta V_B = 2 : 1, answer (c).
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Q40JEE Main 2023 · 24 JAN · SHIFT 1Medium

Thermometry & Thermal Expansion · Area & volume expansion

A hole is drilled in a metal sheet. At 27∘C27^\circ\text{C}, the diameter of hole is 5 cm. When the sheet is heated to 177∘C177^\circ\text{C}, the change in the diameter of hole is d×10−3\text{d} \times 10^{-3} cm. The value of d will be __________ if coefficient of linear expansion of the metal is 1.6×10−5/∘1.6 \times 10^{-5}/^\circC.

Numerical answer type. Work it out, then open the answer.

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Answer: 12

  1. A hole expands like the material around it: Δd=dαΔT\Delta d = d\alpha\Delta T.
  2. Δd=5×1.6×10−5×150=0.012\Delta d = 5 \times 1.6\times10^{-5} \times 150 = 0.012 cm
  3. 0.012=12×10−30.012 = 12\times10^{-3} cm, so d=12d = 12.
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Q41JEE Main 2023 · 8 APR · SHIFT 2Medium

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A steel rod of length 1 m1\text{ m} and cross sectional area 10−4 m210^{-4}\text{ m}^2 is heated from 0∘C0^\circ\text{C} to 200∘C200^\circ\text{C} without being allowed to extend or bend. The compressive tension produced in the rod is ___________ ×104 N\times 10^4\text{ N}. (Given Young's modulus of steel =2×1011 N m−2= 2 \times 10^{11}\text{ N m}^{-2}, coefficient of linear expansion =10−5 K−1= 10^{-5}\text{ K}^{-1})

Numerical answer type. Work it out, then open the answer.

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Answer: 4

  1. F=YAαΔTF = YA\alpha\Delta T
  2. F=2×1011×10−4×10−5×200=4×104F = 2\times10^{11} \times 10^{-4} \times 10^{-5} \times 200 = 4\times10^4 N
  3. So the answer is 4.
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Q42JEE Main 2022 · 30 JUN · SHIFT 1Medium

Thermometry & Thermal Expansion · Temperature scales & thermometers

The pressure of the gas in a constant volume gas thermometer is 100 cm of mercury when placed in melting ice at 1 atm. When the bulb is placed in a liquid, the pressure becomes 180 cm of mercury. Temperature of the liquid is:

(Given 0∘^\circ C = 273 K)

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Answer: (d)

  1. At constant volume, PT=\frac{P}{T}= constant: T273=180100\frac{T}{273}=\frac{180}{100}
  2. T=273×1.8=491.4T=273\times1.8=491.4 K
  3. So the answer is (d) 491 K.
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Q43JEE Main 2022 · 29 JUN · SHIFT 2Medium

Thermometry & Thermal Expansion · Linear expansion

At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm? Both the diameters have been measured at room temperature (27∘^\circC).

(Given: coefficient of linear thermal expansion of gold α\alphaL = 1.4 ×\times 10−-5 K−1{}^{-1})

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Answer: (d)

  1. Ring diameter must grow by ΔD=6.241−6.230=0.011\Delta D = 6.241 - 6.230 = 0.011 cm.
  2. ΔT=ΔDDα=0.0116.230×1.4×10−5≈126\Delta T = \dfrac{\Delta D}{D\alpha} = \dfrac{0.011}{6.230 \times 1.4\times10^{-5}} \approx 126 K
  3. T=27+126≈153∘CT = 27 + 126 \approx 153^\circ\text{C}
  4. So the answer is (d) 152.7∘C152.7^\circ\text{C}.
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Q44JEE Main 2022 · 25 JUN · SHIFT 2Medium

Thermometry & Thermal Expansion · Area & volume expansion

A solid metallic cube having total surface area 24 m2 is uniformly heated. If its temperature is increased by 10∘^\circC, calculate the increase in volume of the cube. (Given α\alpha = 5.0 ×\times 10 −-4 ∘C−^\circ\text{C}^{-}1).

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Answer: (b)

  1. 6a2=24⇒a=26a^2 = 24 \Rightarrow a = 2 m, V=8 m3V = 8\text{ m}^3.
  2. ΔV=V(3α)ΔT=8×3×5×10−4×10=0.12 m3\Delta V = V(3\alpha)\Delta T = 8 \times 3 \times 5\times10^{-4} \times 10 = 0.12\text{ m}^3
  3. 0.12 m3=1.2×105 cm30.12\text{ m}^3 = 1.2\times10^5\text{ cm}^3, answer (b).
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Q45JEE Main 2022 · 25 JUL · SHIFT 1Medium

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A unit scale is to be prepared whose length does not change with temperature and remains 20 cm20\text{ cm}, using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that difference between their lengths remains constant. If length of brass is 40 cm40\text{ cm}, then length of iron will be __________ cm⋅(α\text{cm} \cdot \left(\alpha\right.iron =1.2×10−5 K−1= 1.2 \times 10^{-5}\text{ K}^{-1} and α\alphabrass =1.8×10−5 K−1)= 1.8 \times 10^{-5}\text{ K}^{-1}\Big).

Numerical answer type. Work it out, then open the answer.

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Answer: 60

  1. Difference stays constant if both expand equally: Lbαb=LiαiL_b\alpha_b = L_i\alpha_i.
  2. Li=40×1.8×10−51.2×10−5=60L_i = \dfrac{40 \times 1.8\times10^{-5}}{1.2\times10^{-5}} = 60 cm
  3. Check: 60−40=2060 - 40 = 20 cm. Answer: 60 cm.
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Q46MHT CET 2021 · 21 SEP · SHIFT 1Easy

Thermometry & Thermal Expansion · Temperature scales & thermometers

Find the value of −- 197∘^\circC temperature in Kelvin.

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Answer: (b)

  1. T(K)=t(∘C)+273T(\text{K})=t(^\circ\text{C})+273
  2. T=−197+273=76T=-197+273=76 K
  3. So the answer is (b) 76 K.
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Q47MHT CET 2021 · 20 SEP · SHIFT 1Easy

Thermometry & Thermal Expansion · Temperature scales & thermometers

On an imaginary linear scale of temperature (called 'W' scale) the freezing and boiling points of water are 39∘^\circ W and 239∘^\circ W respectively. The temperature on the new scale corresponding to 39 ∘^\circC temperature on Celsius scale will be

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Answer: (c)

  1. W−39239−39=C−0100\frac{W-39}{239-39}=\frac{C-0}{100}, so W=39+2CW=39+2C.
  2. W=39+2×39=117W=39+2\times 39=117
  3. So the answer is (c) 117∘117^\circW.
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Q48JEE Main 2021 · 16 MAR · SHIFT 2Easy

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A bimetallic strip consists of metals A and B. It is mounted rigidly as shown. The metal A has higher coefficient of expansion compared to that of metal B. When the bimetallic strip is placed in a cold bath, it will:

Figure

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Answer: (b)

  1. In a cold bath both strips shrink, but A (larger α\alpha) shrinks more.
  2. The shorter strip lies on the inside of the curve, so the strip bends towards A.
  3. A is on the left, so the answer is (b) bend towards the left.
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Q49JEE Main 2021 · 25 FEB · SHIFT 1Easy

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: When a rod lying freely is heated, no thermal stress is developed in it. Reason R: On heating, the length of the rod increases.

In the light of the above statements, choose the correct answer from the options given below:

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Answer: (c)

  1. A free rod can expand without restriction, so no thermal stress develops: A is true.
  2. On heating the length increases: R is true.
  3. Stress arises only when expansion is prevented; the mere increase in length does not explain the absence of stress.
  4. So the answer is (c) both true, R not the correct explanation.
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Q50MHT CET 2021 · 23 SEP · SHIFT 1Medium

Thermometry & Thermal Expansion · Linear expansion

Two rods of different metals have coefficients of linear expansion 'α1\alpha_1' and 'α2\alpha_2' respectvely. Their respective lengths are 'L1\text{L}_1' and 'L2\text{L}_2'. At all temperatures (L2−L1\text{L}_2 - \text{L}_1) is same. The correct relation is

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Answer: (d)

  1. (L2−L1)(L_2-L_1) constant means both rods expand by the same amount: L1α1ΔT=L2α2ΔTL_1\alpha_1\Delta T=L_2\alpha_2\Delta T.
  2. So the answer is (d) L1α1=L2α2L_1\alpha_1=L_2\alpha_2.
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Q51MHT CET 2021 · 22 SEP · SHIFT 1Medium

Thermometry & Thermal Expansion · Area & volume expansion

Let 'W1\text{W}_1' be the work done in blowing a soap bubble of radius 'rr' from soap at room temperature. The soap is now heated and second soap bubble of radius '2r2r' is blown from the heated soap solution. If 'W2W_2' is the work done in forming this bubble then

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Answer: (d)

  1. Work to blow a bubble W=8πr2TW = 8\pi r^2 T (two surfaces).
  2. At radius 2r2r with the same TT it would be 4W14W_1.
  3. Surface tension decreases on heating, so T′<TT' < T.
  4. So W2<4W1W_2 < 4W_1, answer (d).
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Q52JEE Main 2021 · 24 FEB · SHIFT 1Medium

Thermometry & Thermal Expansion · Area & volume expansion

Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is 'α\alpha'. The metal sheet is heated uniformly, by a small temperature Δ\DeltaT, so that its new temperature is T + Δ\DeltaT. Calculate the increase in the volume of the metal box.

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Answer: (a)

  1. The box expands like a solid cube of the same metal: γ=3α\gamma = 3\alpha.
  2. ΔV=a3(3α)ΔT\Delta V = a^3(3\alpha)\Delta T
  3. So the answer is (a) 3a3αΔT3a^3\alpha\Delta T.
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Q53JEE Main 2021 · 22 JUL · SHIFT 2Difficult

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

The area of cross-section of a railway track is 0.01 m2. The temperature variation is 10∘^{\circ}C. Coefficient of linear expansion of material of track is 10−5/∘^{-5}/^{\circ} C. The energy stored per meter in the track is ____________ J/m.

(Young's modulus of material of track is 1011 Nm−^{-}2)

Numerical answer type. Work it out, then open the answer.

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Answer: 05

  1. Thermal stress σ=YαΔT=1011×10−5×10=107\sigma = Y\alpha\Delta T = 10^{11} \times 10^{-5} \times 10 = 10^7 Pa.
  2. Energy per unit length =12σ2YA=12×10141011×0.01= \dfrac{1}{2}\dfrac{\sigma^2}{Y}A = \dfrac{1}{2} \times \dfrac{10^{14}}{10^{11}} \times 0.01
  3. =5= 5 J/m
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Q54JEE Main 2020 · 7 JAN · SHIFT 1Easy

Thermometry & Thermal Expansion · Area & volume expansion

A non-isotropic solid metal cube has coefficients of linear expansion as:

5 ×\times 10-5/oC along the x-axis and 5 ×\times 10-6/oC along the y and the z-axis. If the coefficient of volume expansion of the solid is C ×\times 10-6/oC then the value of C is

Numerical answer type. Work it out, then open the answer.

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Answer: 60

  1. For a cube, γ=αx+αy+αz\gamma=\alpha_x+\alpha_y+\alpha_z.
  2. γ=50×10−6+5×10−6+5×10−6=60×10−6/∘\gamma=50\times10^{-6}+5\times10^{-6}+5\times10^{-6}=60\times10^{-6}/^\circC
  3. So C=60C=60.
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Q55JEE Main 2020 · 5 SEP · SHIFT 2Medium

Thermometry & Thermal Expansion · Linear expansion

Two different wires having lengths L1 and L2, and respective temperature coefficient of linear expansion α\alpha1 and α\alpha2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is:

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Answer: (d)

  1. Total expansion: ΔL=(L1α1+L2α2)ΔT\Delta L = (L_1\alpha_1 + L_2\alpha_2)\Delta T.
  2. Also ΔL=(L1+L2)αeffΔT\Delta L = (L_1+L_2)\alpha_{eff}\Delta T.
  3. αeff=α1L1+α2L2L1+L2\alpha_{eff} = \dfrac{\alpha_1 L_1 + \alpha_2 L_2}{L_1 + L_2}, so the answer is (d).
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Q56JEE Main 2020 · 2 SEP · SHIFT 2Medium

Thermometry & Thermal Expansion · Area & volume expansion

When the temperature of a metal wire is increased from 0oC to 10oC, its length increases by 0.02%. The percentage change in its mass density will be closest to:

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Answer: (b)

  1. Linear strain =0.02%= 0.02\%, so volume increases by 3×0.02=0.06%3 \times 0.02 = 0.06\%.
  2. Mass is constant, so ρ∝1/V\rho \propto 1/V: density falls by about 0.06%0.06\%.
  3. So the answer is (b) 0.06.
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Q57JEE Main 2020 · 3 SEP · SHIFT 1Difficult

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A bakelite beaker has volume capacity of 500 cc at 30oC. When it is partially filled with Vm volume (at 30oC) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If γ\gamma(beaker) = 6 × 10–6oC–1 and γ\gamma(mercury) = 1.5 × 10–4oC–1, where γ\gamma is the coefficient of volume expansion, then Vm (in cc) is close to ____.

Numerical answer type. Work it out, then open the answer.

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Answer: 20

  1. Unfilled volume constant: Vmγm=VbγbV_m\gamma_m = V_b\gamma_b.
  2. Vm=500×6×10−61.5×10−4=20V_m = \dfrac{500 \times 6\times10^{-6}}{1.5\times10^{-4}} = 20 cc
  3. Answer: 20 cc.
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Q58JEE Main 2019 · 11 JAN · SHIFT 2Medium

Thermometry & Thermal Expansion · Temperature scales & thermometers

A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and x0/3 when in contact with ice. What is the temperature of an object inoC, if this thermometer in the contact with the object reads x0/2?

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Answer: (c)

  1. t100=x−x0/3x0−x0/3\frac{t}{100}=\frac{x-x_0/3}{x_0-x_0/3}
  2. t100=x0/2−x0/32x0/3=1/62/3=14\frac{t}{100}=\frac{x_0/2-x_0/3}{2x_0/3}=\frac{1/6}{2/3}=\frac{1}{4}
  3. So t=25∘t=25^\circC, answer (c).
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Q59NEET 2019Medium

Thermometry & Thermal Expansion · Linear expansion

A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is: (α\alphaCu = 1.7 × 10–5 K–1 and α\alphaAl = 2.2 × 10–5 K–1)

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Answer: (c)

  1. The difference in length stays constant when expansions are equal: LCuαCu=LAlαAlL_{Cu}\alpha_{Cu}=L_{Al}\alpha_{Al}.
  2. LAl=88×1.72.2=68L_{Al}=\frac{88\times1.7}{2.2}=68 cm
  3. So the answer is (c) 68 cm.
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Q60JEE Main 2019 · 11 JAN · SHIFT 2Medium

Thermometry & Thermal Expansion · Linear expansion

Two rods A and B of identical dimensions are at temperature 30°C. If A is heated upto 180oC and B upto ToC, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4: 3, then the value of T is

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Answer: (c)

  1. Same initial length and same final length means equal expansion: αA(180−30)=αB(T−30)\alpha_A (180-30) = \alpha_B (T-30).
  2. T−30=150×43=200T - 30 = 150 \times \dfrac{4}{3} = 200
  3. So T=230∘CT = 230^\circ\text{C}, answer (c).
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Q61JEE Main 2019 · 9 JAN · SHIFT 1Medium

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion α\alpha/oC. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises by Δ\DeltaT K. Young's modulus, Y, for this metal is:

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Answer: (a)

  1. Prevented expansion gives thermal strain αΔT\alpha\Delta T.
  2. FA=YαΔT\dfrac{F}{A} = Y\alpha\Delta T (a change of ΔT\Delta T is the same in K and ∘^\circC).
  3. Y=FAαΔTY = \dfrac{F}{A\alpha\Delta T}, answer (a).
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Q62MHT CET 2019 · 2 MAY · SHIFT 1Difficult

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A clock pendulum having coefficient of linear expansion. α=9×10−7/∘C−1\alpha = 9 \times 10^{-7}/{}^{\circ}\text{C}^{-1} has a period of 0.5 s at 20∘C20^{\circ}\text{C}. If the clock is used in a climate, where the temperature is 30∘C30^{\circ}\text{C}, how much time does the clock lose in each oscillation? ( g=g = constant)

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Answer: (d)

  1. ΔTT=12αΔθ\dfrac{\Delta T}{T} = \dfrac{1}{2}\alpha\Delta\theta
  2. ΔT=12×9×10−7×10×0.5=2.25×10−6\Delta T = \dfrac{1}{2} \times 9\times10^{-7} \times 10 \times 0.5 = 2.25\times10^{-6} s
  3. So the answer is (d).
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Q63JEE Main 2019 · 12 APR · SHIFT 1Difficult

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

At 40o C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40oC to 20oC it regains its original length of 0.2 m. The value of M is close to: (Coefficient of linear expansion and Young’s modulus of brass are 10–5 /oC and 1011 N/m 2, respectively; g= 10 ms–2)

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Answer: (c)

  1. Stretch by the load equals thermal contraction: Mgπr2Y=αΔθ\dfrac{Mg}{\pi r^2 Y} = \alpha\Delta\theta.
  2. M=Yπr2αΔθg=1011×π×10−6×10−5×2010M = \dfrac{Y\pi r^2\alpha\Delta\theta}{g} = \dfrac{10^{11} \times \pi \times 10^{-6} \times 10^{-5} \times 20}{10}
  3. M=2π≈6.3M = 2\pi \approx 6.3 kg; the closest option is 9 kg.
  4. So the answer is (c) 9 kg.
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Q64JEE Main 2017 · 9 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A steel rail of length 5 m and area of cross section 40cm2 is prevented from expanding along its length while the temperature rises by 10oC. If coefficient of linear expansion and Young’s modulus of steel are 1.2×10−5 K−1 and 2×1011 Nm−2 respectively, the force developed in the rail is approximately:

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Answer: (b)

  1. Thermal force F=YAαΔTF = YA\alpha\Delta T.
  2. F=2×1011×40×10−4×1.2×10−5×10=9.6×104F = 2\times10^{11} \times 40\times10^{-4} \times 1.2\times10^{-5} \times 10 = 9.6\times10^4 N
  3. So F≈1×105F \approx 1\times10^5 N, answer (b).
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Q65JEE Main 2017 · 8 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by Δ\DeltaT. The net change in its length is zero. Let ℓ\ell be the length of the rod, A its area of cross-section,Y its Young’s modulus, and α\alpha its coefficient of linear expansion. Then, F is equal to:

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Answer: (c)

  1. Compression must cancel thermal strain: FAY=αΔT\dfrac{F}{AY} = \alpha\Delta T.
  2. So F=AYαΔTF = AY\alpha\Delta T, answer (c).
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Q66JEE Main 2017Difficult

Thermometry & Thermal Expansion · Area & volume expansion

An external pressure P is applied on a cube at 0oC so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α\alpha is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

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Answer: (a)

  1. Compression: ΔVV=PK\dfrac{\Delta V}{V} = \dfrac{P}{K}.
  2. Heating must give the same ΔVV=3αΔT\dfrac{\Delta V}{V} = 3\alpha\Delta T.
  3. ΔT=P3αK\Delta T = \dfrac{P}{3\alpha K}, answer (a).
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Q67JEE Main 2016 · 9 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period 2 s at T=0oC. If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is α\alpha then the value of S is:

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Answer: (a)

  1. T=2πL/gT = 2\pi\sqrt{L/g}, so ΔTT=12αθ\dfrac{\Delta T}{T} = \dfrac{1}{2}\alpha\theta.
  2. ΔT=22αθ=αθ\Delta T = \dfrac{2}{2}\alpha\theta = \alpha\theta (with T=2T = 2 s).
  3. Slope S=αS = \alpha, answer (a).
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Q68JEE Main 2016Difficult

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A pendulum clock loses 12 s12\text{ s} a day if the temperature is 40∘C40^{\circ}\text{C} and gains 4 s4\text{ s} a day if the temperature is 20∘C.20^{\circ}\text{C}. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α)(\alpha) of the metal of the pendulum shaft are respectively:

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Answer: (c)

  1. Time lost or gained per day: Δt=12α(θ−θ0)×86400\Delta t = \dfrac{1}{2}\alpha(\theta - \theta_0) \times 86400.
  2. 12=12α(40−θ0)(86400)12 = \tfrac{1}{2}\alpha(40-\theta_0)(86400) and 4=12α(θ0−20)(86400)4 = \tfrac{1}{2}\alpha(\theta_0-20)(86400)
  3. Dividing: 3(θ0−20)=40−θ0⇒θ0=25∘C3(\theta_0 - 20) = 40 - \theta_0 \Rightarrow \theta_0 = 25^\circ\text{C}
  4. Adding: 16=12α(20)(86400)⇒α≈1.85×10−5/∘C16 = \tfrac{1}{2}\alpha(20)(86400) \Rightarrow \alpha \approx 1.85\times10^{-5}/^\circ\text{C}
  5. So the answer is (c).
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Q69JEE Main 2013 · 9 APR · SHIFT 1Easy

Thermometry & Thermal Expansion · Temperature scales & thermometers

On a linear temperature scale Y, water freezes at −160∘-160^\circ Y and boils at −50∘-50^\circ Y. On this Y scale, a temperature of 340 K would be read as: (water freezes at 273 K and boils at 373 K)

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Answer: (c)

  1. 340340 K =67∘=67^\circC.
  2. Y−(−160)−50−(−160)=67100⇒Y+160=110×0.67=73.7\frac{Y-(-160)}{-50-(-160)}=\frac{67}{100}\Rightarrow Y+160=110\times 0.67=73.7
  3. Y=−86.3∘Y=-86.3^\circY
  4. So the answer is (c) −86.3∘-86.3^\circY.
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Q70JEE Main 2013 · 23 APR · SHIFT 1Medium

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

The ratio of the coefficient of volume expansion of a glass container to that of a viscous liquid kept inside the container is 1: 4. What fraction of the inner volume of the container should the liquid occupy so that the volume of the remaining vacant space will be same at all temperatures?

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Answer: (b)

  1. Vacant space stays constant if liquid and container expand by the same volume: Vlγl=VcγcV_l\gamma_l = V_c\gamma_c.
  2. VlVc=γcγl=14\dfrac{V_l}{V_c} = \dfrac{\gamma_c}{\gamma_l} = \dfrac{1}{4}
  3. So the answer is (b) 1 : 4.
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Q71JEE Main 2012 · AIEEEDifficult

Thermometry & Thermal Expansion · Thermal stress, pendulum & liquids

A wooden wheel of radius RR is made of two semicircular part (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area SS and length L. LL.\ L is slightly less than 2πR.2\pi R. To fit the ring on the wheel, it is heated so that its temperature rises by ΔT\Delta T and it just steps over the wheel. As it cools down to surrounding temperature, it process the semicircular parts together. If the coefficient of linear expansion of the metal is α\alpha, and its Young's modulus is Y,Y, the force that one part of the wheel applies on the other part is:

Figure

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Answer: (d)

TTring half (top)net pull = 2T
  1. Ring strain after cooling: 2πR−LL=αΔT\dfrac{2\pi R - L}{L} = \alpha\Delta T.
  2. Tension in the strip: T=YSαΔTT = YS\alpha\Delta T.
  3. Each half of the ring pulls the two halves together with TT at both cut ends, so force =2T= 2T.
  4. F=2SYαΔTF = 2SY\alpha\Delta T, answer (d).
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Q72NEET 1999 · AIPMTMedium

Thermometry & Thermal Expansion · Linear expansion

The coefficients of linear expansions of brass and steel are α1\alpha_1 and α2\alpha_2 respectively. When we take a brass rod of length l1l_1 and a steel rod of length l2l_2 at 0∘C0^\circ\text{C}, then the difference in their lengths (l2−l1)(l_2 - l_1) will remain the same at all temperatures if

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Answer: (a)

  1. At temperature θ\theta: l2′−l1′=(l2−l1)+(l2α2−l1α1)θl_2' - l_1' = (l_2 - l_1) + (l_2\alpha_2 - l_1\alpha_1)\theta.
  2. The difference stays constant only if the two expansions are equal: l1α1=l2α2l_1\alpha_1 = l_2\alpha_2.
  3. So the answer is (a) α1l1=α2l2\alpha_1 l_1 = \alpha_2 l_2.
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Q73NEET 1992 · AIPMTEasy

Thermometry & Thermal Expansion · Temperature scales & thermometers

Mercury thermometer can be used to measure temperature upto

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Answer: (c)

  1. Mercury freezes at about −39∘-39^\circC and boils at about 357∘357^\circC.
  2. So a mercury thermometer works up to about 360∘360^\circC.
  3. So the answer is (c) 360∘360^\circC.
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Q74NEET 1990 · AIPMTEasy

Thermometry & Thermal Expansion · Temperature scales & thermometers

A Centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers 140∘F140^\circ\text{F}. What is the fall in temperature as registered by the centigrade thermometer?

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Answer: (c)

  1. Boiling water: 100∘100^\circC. At 140∘140^\circF: C=59(140−32)=60∘C=\frac{5}{9}(140-32)=60^\circC.
  2. Fall =100−60=40∘=100-60=40^\circC
  3. So the answer is (c) 40∘40^\circC.
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Q75JEE Main 2026 · 28 JAN · SHIFT 1Easy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

Which of the following best represents the temperature versus heat supplied graph for water, in the range of −20∘C-20^{\circ}\text{C} to 120∘C120^{\circ}\text{C}?

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Answer: (c)

0100meltingboilingHeatT (°C)
  1. Ice warms from −20∘C-20^{\circ}\text{C} to 0∘C0^{\circ}\text{C}: temperature rises.
  2. At 0∘C0^{\circ}\text{C} ice melts: flat part (latent heat of fusion).
  3. Water warms from 0∘C0^{\circ}\text{C} to 100∘C100^{\circ}\text{C}: rises, then flat at 100∘C100^{\circ}\text{C} while it boils.
  4. Steam then warms to 120∘C120^{\circ}\text{C}: rises again. Only graph (c) has flat parts at both 0∘C0^{\circ}\text{C} and 100∘C100^{\circ}\text{C}.
  5. So the answer is (c).
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Q76MHT CET 2026 · 19 MAY · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

A lead bullet moving with velocity 'vv' strikes a wall and stops. 75%75\% of its energy is converted into heat. If the increase in the temperature is v2nJs\frac{v^2}{nJs}, then the value of nn is

{s=\{s = specific heat of lead,J=, J = mechanical equivalent of heat}\}

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Answer: (a)

  1. 0.75×12mv2=J m s ΔT0.75 \times \tfrac{1}{2}mv^2 = J\,m\,s\,\Delta T
  2. ΔT=3v28Js=v283Js\Delta T = \dfrac{3v^2}{8Js} = \dfrac{v^2}{\tfrac{8}{3}Js}
  3. So n=83n = \tfrac{8}{3}, answer (a).
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Q77JEE Main 2026 · 21 JAN · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27∘C27^{\circ}\text{C} to 87∘C87^{\circ}\text{C}. The rate of consumption of the gas is ‾ g/s\underline{\hspace{3em}}\text{ g/s}.

(Take heat of combustion of gas =5.0×104 J/g= 5.0 \times 10^{4}\text{ J/g}, specific heat capacity of water =4200 J/kg⋅∘C= 4200\text{ J/kg}\cdot{}^{\circ}\text{C})

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Answer: (d)

  1. Mass flow =5 kg/min=112 kg/s= 5\text{ kg/min} = \tfrac{1}{12}\text{ kg/s}.
  2. Heat per second =112×4200×60=21000 J/s= \tfrac{1}{12} \times 4200 \times 60 = 21000\text{ J/s}
  3. Gas rate =210005×104=0.42 g/s= \dfrac{21000}{5 \times 10^4} = 0.42\text{ g/s}
  4. So the answer is (d) 0.420.42.
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Q78MHT CET 2026 · 14 MAY · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

11 g of water of volume 11 cm3^{3} at 100∘100^{\circ}C is converted into steam of same temperature under normal atmospheric pressure (10510^{5} Pa). The volume of steam formed equals 16711671 cm3^{3}. If the latent heat of vaporisation of water is 22562256 J/g, the change in internal energy is

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Answer: (b)

  1. First law: ΔU=Q−W\Delta U = Q - W, with Q=mL=2256 JQ = mL = 2256\text{ J}.
  2. W=PΔV=105×(1671−1)×10−6=167 JW = P\Delta V = 10^5 \times (1671 - 1) \times 10^{-6} = 167\text{ J}
  3. ΔU=2256−167=2089 J\Delta U = 2256 - 167 = 2089\text{ J}
  4. So the answer is (b) 2089 J2089\text{ J}.
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Q79JEE Main 2026 · 5 APR · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

The heat extracted out of xx gram of water initially at 50∘C50^{\circ}\text{C} to cool it down to 0∘C0^{\circ}\text{C} is sufficient to evaporate (1000−x)(1000 - x) gram of water also initially at 50∘C50^{\circ}\text{C}. The value of xx (closest integer) is ______. (Take latent heat of water 2256 kJ/kg2256\text{ kJ/kg}, specific heat capacity of water 4200 J/kg.K4200\text{ J/kg.K})

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 922

  1. Heat from xx g cooling 50→0∘C50 \to 0^{\circ}\text{C}: x×4.2×50=210x Jx \times 4.2 \times 50 = 210x\text{ J}.
  2. Heat for (1000−x)(1000 - x) g: warm to 100∘C100^{\circ}\text{C} and evaporate: (1000−x)(210+2256)=2466(1000−x) J(1000 - x)(210 + 2256) = 2466(1000 - x)\text{ J}.
  3. 210x=2466(1000−x)⇒2676x=2466000210x = 2466(1000 - x) \Rightarrow 2676x = 2466000
  4. x≈921.5≈922x \approx 921.5 \approx 922
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Q80MHT CET 2026 · 20 MAY · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

540540 gram of ice at 0∘0^{\circ} C is mixed with 540540 gram of water at 90∘90^{\circ} C. The final temperature of the mixture will be [Latent heat of fusion of water = 80 cal/g]

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Answer: (c)

  1. Heat from water cooling to 0∘C0^{\circ}\text{C}: 540×90=48600 cal540 \times 90 = 48600\text{ cal}; to melt ice: 540×80=43200 cal540 \times 80 = 43200\text{ cal}.
  2. All ice melts; extra 5400 cal5400\text{ cal} heats 1080 g1080\text{ g} of water.
  3. T=54001080=5∘CT = \dfrac{5400}{1080} = 5^{\circ}\text{C}
  4. So the answer is (c) 5∘C5^{\circ}\text{C}.
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Q81MHT CET 2026 · 20 APR · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

A piece of metal of 850 K is dropped into 1 kg of water at 300 K. If equilibrium temperature of water is 350 K then the heat capacity of the metal expressed in JK\frac{J}{K} is (\Bigg(Sp. heat of water =1 calg ∘C)= 1\ \frac{\text{cal}}{\text{g}\,{}^{\circ}\text{C}}\Bigg)

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Answer: (a)

  1. Heat gained by water =1000×1×50=50000 cal=2.1×105 J= 1000 \times 1 \times 50 = 50000\text{ cal} = 2.1 \times 10^5\text{ J}
  2. Metal falls by 850−350=500 K850 - 350 = 500\text{ K}: C×500=2.1×105C \times 500 = 2.1 \times 10^5
  3. C=420 J/KC = 420\text{ J/K}
  4. So the answer is (a) 420.
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Q82MHT CET 2026 · 19 APR · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

1515 g of ice at 0∘0^{\circ}C is added to a vessel containing water at 40∘40^{\circ}C. The mass of water and water equivalent of the vessel is 6060 g. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be

[ LLice =80= 80 cal/g, SSwater =1= 1 cal/g ]

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Answer: (c)

  1. Heat from 60 g cooling to 0∘C0^{\circ}\text{C}: 60×40=2400 cal60 \times 40 = 2400\text{ cal}; to melt ice: 15×80=1200 cal15 \times 80 = 1200\text{ cal}, so all ice melts.
  2. 60(40−T)=1200+15T60(40 - T) = 1200 + 15T
  3. 2400−60T=1200+15T⇒T=16∘C2400 - 60T = 1200 + 15T \Rightarrow T = 16^{\circ}\text{C}
  4. So the answer is (c) 16∘C16^{\circ}\text{C}.
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Q83JEE Main 2026 · 28 JAN · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

10 kg of ice at −10∘C-10^{\circ}\text{C} is added to 100 kg of water to lower its temperature from 25 ∘C^{\circ}\text{C}. Consider no heat exchange to surroundings. The decrement to the temperature of water is ‾ ∘C\underline{\hspace{3em}}\ {}^{\circ}\text{C}. (specific heat of ice =2100 J/kg.∘C= 2100\text{ J/kg.}{}^{\circ}\text{C}, specific heat of water =4200 J/kg.∘C= 4200\text{ J/kg.}{}^{\circ}\text{C}, latent heat of fusion of ice =3.36×105 J/kg= 3.36 \times 10^{5}\text{ J/kg})

Show answer and solution

Answer: (b)

  1. Ice gains: 10×2100×10+10×3.36×105+10×4200 T=3.57×106+42000 T10 \times 2100 \times 10 + 10 \times 3.36 \times 10^5 + 10 \times 4200\,T = 3.57 \times 10^6 + 42000\,T
  2. Water loses: 100×4200(25−T)=420000(25−T)100 \times 4200 (25 - T) = 420000(25 - T)
  3. 1.05×107−420000T=3.57×106+42000T⇒T=15∘C1.05 \times 10^7 - 420000T = 3.57 \times 10^6 + 42000T \Rightarrow T = 15^{\circ}\text{C}
  4. Decrease =25−15=10∘C= 25 - 15 = 10^{\circ}\text{C}
  5. So the answer is (b) 10.
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Q84JEE Main 2026 · 24 JAN · SHIFT 1Difficult

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

Density of water at 4∘C4^{\circ}\text{C} and 20∘C20^{\circ}\text{C} are 1000 kg/m31000\text{ kg/m}^{3} and 998 kg/m3998\text{ kg/m}^{3} respectively. The increase in internal energy of 4 kg of water when it is heated from 4∘C4^{\circ}\text{C} to 20∘C20^{\circ}\text{C} is ______ J.

(specific heat capacity of water =4.2 kJ/kg K= 4.2\text{ kJ/kg K} and 1 atmospheric pressure =105 Pa= 10^{5}\text{ Pa})

Show answer and solution

Answer: (a)

  1. First law: ΔU=Q−PΔV\Delta U = Q - P\Delta V.
  2. Q=mcΔT=4×4200×16=268800 JQ = mc\Delta T = 4 \times 4200 \times 16 = 268800\ \text{J}
  3. ΔV=4998−41000≈8.016×10−6 m3\Delta V = \dfrac{4}{998} - \dfrac{4}{1000} \approx 8.016 \times 10^{-6}\ \text{m}^3, so PΔV=105×8.016×10−6≈0.8 JP\Delta V = 10^5 \times 8.016 \times 10^{-6} \approx 0.8\ \text{J}
  4. ΔU=268800−0.8=268799.2 J\Delta U = 268800 - 0.8 = 268799.2\ \text{J}
  5. So the answer is (a) 268799.2.
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Q85JEE Main 2025 · 8 APR · SHIFT 2Easy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool.

(Take g = 10 m/s2, specific heat of water = 4200 J/(kg K))

Show answer and solution

Answer: (c)

  1. All potential energy becomes heat: mgh=mcΔTmgh = mc\Delta T.
  2. ΔT=ghc=10×2004200≈0.48\Delta T = \dfrac{gh}{c} = \dfrac{10 \times 200}{4200} \approx 0.48 K
  3. So the answer is (c).
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Q86JEE Main 2025 · 2 APR · SHIFT 2Easy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

Match List - I with List - II.

List - I List - II\text{List - II}

(A) Heat capacity of body (I) J kg−1\text{J kg}^{-1}

(B) Specific heat capacity of body (II) J K−1\text{J K}^{-1}

(C) Latent heat (III) J kg−1 K−1\text{J kg}^{-1}\text{ K}^{-1}

(D) Thermal conductivity (IV) J m−1 K−1 s−1\text{J m}^{-1}\text{ K}^{-1}\text{ s}^{-1}

Choose the correct answer from the options given below:

Show answer and solution

Answer: (d)

  1. (A) Heat capacity =Q/ΔT⇒J K−1= Q/\Delta T \Rightarrow \text{J K}^{-1} (II).
  2. (B) Specific heat =Q/(mΔT)⇒J kg−1K−1= Q/(m\Delta T) \Rightarrow \text{J kg}^{-1}\text{K}^{-1} (III).
  3. (C) Latent heat =Q/m⇒J kg−1= Q/m \Rightarrow \text{J kg}^{-1} (I).
  4. (D) Thermal conductivity K=Q xA ΔT t⇒J m−1K−1s−1K = \dfrac{Q\,x}{A\,\Delta T\,t} \Rightarrow \text{J m}^{-1}\text{K}^{-1}\text{s}^{-1} (IV).
  5. So the answer is (d) (A)-(II), (B)-(III), (C)-(I), (D)-(IV).
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Q87JEE Main 2025 · 23 JAN · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J, then the mass of the bullet is _______ grams. (Latent heat of fusion of lead =2.5×104 J kg−1= 2.5 \times 10^{4}\text{ J kg}^{-1} and specific heat capacity of lead =125 J kg−1 K−1)= 125\text{ J kg}^{-1}\text{ K}^{-1}\Big)

Show answer and solution

Answer: (d)

  1. Heat =m(cΔT+L)=m(125×300+2.5×104)=62500 m= m(c\Delta T + L) = m(125 \times 300 + 2.5 \times 10^4) = 62500\,m
  2. 62500 m=625⇒m=0.01 kg=10 g62500\,m = 625 \Rightarrow m = 0.01\text{ kg} = 10\text{ g}
  3. So the answer is (d) 10.
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Q88JEE Main 2025 · 22 JAN · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

An amount of ice of mass 10−3 kg10^{-3}\text{ kg} and temperature −10∘C-10^{\circ}\text{C} is transformed to vapour of temperature 110∘C110^{\circ}\text{C} by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice =2100 J kg−1 K−1= 2100\text{ J kg}^{-1}\text{ K}^{-1}, specific heat of water =4180 J kg−1 K−1= 4180\text{ J kg}^{-1}\text{ K}^{-1}, specific heat of steam =1920 J kg−1 K−1= 1920\text{ J kg}^{-1}\text{ K}^{-1}, Latent heat of ice =3.35×105 J kg−1= 3.35 \times 10^{5}\text{ J kg}^{-1} and Latent heat of steam =2.25×106 J kg−1= 2.25 \times 10^{6}\text{ J kg}^{-1})

Show answer and solution

Answer: (b)

  1. Ice −10→0∘C-10 \to 0^{\circ}\text{C}: 10−3×2100×10=21 J10^{-3} \times 2100 \times 10 = 21\text{ J}; melting: 335 J335\text{ J}.
  2. Water 0→100∘C0 \to 100^{\circ}\text{C}: 10−3×4180×100=418 J10^{-3} \times 4180 \times 100 = 418\text{ J}; vaporising: 2250 J2250\text{ J}.
  3. Steam 100→110∘C100 \to 110^{\circ}\text{C}: 10−3×1920×10=19.2 J10^{-3} \times 1920 \times 10 = 19.2\text{ J}
  4. Total =21+335+418+2250+19.2≈3043 J= 21 + 335 + 418 + 2250 + 19.2 \approx 3043\text{ J}
  5. So the answer is (b) 3043 J3043\text{ J}.
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Q89MHT CET 2025 · 23 APR · SHIFT 2Difficult

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

A stationary object at 4∘C4^{\circ}\text{C} and weighing 3.5 kg falls from a height of 2000 m on snow mountain at 0∘C0^{\circ}\text{C}. If the temperature of the object just before hitting the snow is 0∘C0^{\circ}\text{C} and the object comes to rest immediately then the quantity of ice that melts is (Acceleration due to gravity =10 m/s2= 10\text{ m/s}^{2}, Latent heat of ice =3.5×105 J/kg= 3.5 \times 10^{5}\text{ J/kg})

Show answer and solution

Answer: (c)

  1. The object stops at 0∘0^{\circ}C, so its mechanical energy mghmgh goes into melting ice: mgh=miceLmgh = m_{ice}L.
  2. mgh=3.5×10×2000=7×104 Jmgh = 3.5 \times 10 \times 2000 = 7 \times 10^{4}\ \text{J}
  3. mice=7×1043.5×105=0.2 kgm_{ice} = \dfrac{7 \times 10^{4}}{3.5 \times 10^{5}} = 0.2\ \text{kg}
  4. So the answer is (c) 200 gram.
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Q90JEE Main 2024 · 30 JAN · SHIFT 2Easy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

A block of ice at −10∘C-10^{\circ}\text{C} is slowly heated and converted to steam at 100∘C100^{\circ}\text{C}. Which of the following curves represent the phenomenon qualitatively:

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Answer: (a)

  1. Ice at −10∘C-10^{\circ}\text{C}: temperature first rises to 0∘C0^{\circ}\text{C} (sloping line starting from the initial temperature).
  2. Melting at 0∘C0^{\circ}\text{C}: horizontal part.
  3. Water warms to 100∘C100^{\circ}\text{C}: sloping part, then horizontal part while it turns to steam at 100∘C100^{\circ}\text{C}.
  4. (b) and (d) show heat supplied with no temperature change at the start, and (c) has a vertical jump, so they are wrong.
  5. So the answer is (a).
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Q91MHT CET 2024 · 4 MAY · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

A lead bullet moving with velocity ' vv ' strikes a wall and stops. If 50%50\% of its energy is converted into heat, then the increase in temperature is ( s=s = specific heat of lead)

Show answer and solution

Answer: (b)

  1. 0.5×12mv2=J m s ΔT0.5 \times \tfrac{1}{2}mv^2 = J\,m\,s\,\Delta T
  2. ΔT=v24Js\Delta T = \dfrac{v^2}{4Js}
  3. So the answer is (b) v24Js\dfrac{v^2}{4Js}.
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Q92JEE Main 2023 · 31 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

A water heater of power 2000 W2000\text{ W} is used to heat water. The specific heat capacity of water is 4200 J kg−1 K−14200\text{ J kg}^{-1}\text{ K}^{-1}. The efficiency of heater is 70%70\%. Time required to heat 2 kg2\text{ kg} of water from 10∘C10^{\circ}\text{C} to 60∘C60^{\circ}\text{C} is _________ s.

(Assume that the specific heat capacity of water remains constant over the temperature range of the water).

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 300

  1. Useful power =0.7×2000=1400 W= 0.7 \times 2000 = 1400\text{ W}
  2. Heat needed =2×4200×50=4.2×105 J= 2 \times 4200 \times 50 = 4.2 \times 10^5\text{ J}
  3. t=4.2×1051400=300 st = \dfrac{4.2 \times 10^5}{1400} = 300\text{ s}
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Q93JEE Main 2023 · 29 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

Heat energy of 184 kJ is given to ice of mass 600 g at −12∘C-12^{\circ}\text{C}. Specific heat of ice is 2222.3 J kg−1 ∘C−12222.3\text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1} and latent heat of ice is 336 kJ kg−1\text{kJ kg}^{-1}

A. Final temperature of system will be 0∘^{\circ} C.

B. Final temperature of the system will be greater than 0∘^{\circ}C.

C. The final system will have a mixture of ice and water in the ratio of 5: 1.

D. The final system will have a mixture of ice and water in the ratio of 1: 5.

E. The final system will have water only.

Choose the correct answer from the options given below:

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Answer: (d)

  1. Ice −12→0∘C-12 \to 0^{\circ}\text{C}: 0.6×2222.3×12≈16 kJ0.6 \times 2222.3 \times 12 \approx 16\text{ kJ}.
  2. Heat left =184−16=168 kJ= 184 - 16 = 168\text{ kJ}; ice melted =168336=0.5 kg= \dfrac{168}{336} = 0.5\text{ kg}.
  3. Ice left =0.1 kg= 0.1\text{ kg}, water =0.5 kg= 0.5\text{ kg}, temperature stays 0∘C0^{\circ}\text{C}: ice : water =1:5= 1:5.
  4. A and D are true; B, C, E are false.
  5. So the answer is (d) A and D only.
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Q94MHT CET 2023 · 10 MAY · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

A piece of metal at 850 K850\text{ K} is dropped into 1 kg1\text{ kg} water at 300 K300\text{ K}. If the equilibrium temperature of water is 350 K350\text{ K} then the heat capacity of the metal, expressed in J K−1\text{J K}^{-1} is (1 cal=4.2 J)(1\text{ cal} = 4.2\text{ J})

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Answer: (d)

  1. Heat gained by water =1×4200×50=2.1×105 J= 1 \times 4200 \times 50 = 2.1 \times 10^5\text{ J}
  2. Metal falls by 850−350=500 K850 - 350 = 500\text{ K}: C×500=2.1×105C \times 500 = 2.1 \times 10^5
  3. C=420 J K−1C = 420\text{ J K}^{-1}, which is option (a).
  4. The given key marks (d) No Solution.
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Q95MHT CET 2022 · 11 AUG · SHIFT 2Easy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

Heat given to a body, which raises its temperature by 1ºC is known as

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Answer: (b)

  1. Heat needed to raise the temperature of the whole body by 1∘1^\circC is its thermal (heat) capacity, C=msC = ms.
  2. Specific heat is per unit mass, so it does not fit.
  3. So the answer is (b) thermal capacity.
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Q96JEE Main 2022 · 24 JUN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

A 100 g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60 ms−^{-}1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail? [Specific heat capacity of iron = 0.42 Jg−^{-}1 ∘C−^{\circ}\text{C}^{-}1]

Show answer and solution

Answer: (c)

  1. KE of hammer =12(1.5)(60)2=2700 J= \tfrac{1}{2}(1.5)(60)^2 = 2700\text{ J}; heat to nail =14×2700=675 J= \tfrac{1}{4} \times 2700 = 675\text{ J}.
  2. ΔT=Qmc=675100×0.42=16.07∘C\Delta T = \dfrac{Q}{mc} = \dfrac{675}{100 \times 0.42} = 16.07^{\circ}\text{C}
  3. So the answer is (c) 16.07∘C16.07^{\circ}\text{C}.
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Q97JEE Main 2022 · 26 JUN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

A geyser heats water flowing at a rate of 2.0 kg per minute from 30∘^{\circ}C to 70∘^{\circ} C. If geyser operates on a gas burner, the rate of combustion of fuel will be ___________ g min−^{-}1.

[Heat of combustion = 8 ×\times 103 Jg−^{-}1, Specific heat of water = 4.2 Jg−^{-}1 ∘C−^{\circ}\text{C}^{-}1]

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 42

  1. Heat per minute =2000×4.2×40=3.36×105 J/min= 2000 \times 4.2 \times 40 = 3.36 \times 10^5\text{ J/min}
  2. Fuel rate =3.36×1058×103=42 g min−1= \dfrac{3.36 \times 10^5}{8 \times 10^3} = 42\text{ g min}^{-1}
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Q98JEE Main 2022 · 27 JUN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is:

(Given: initial temperature of the bullet = 127∘^{\circ}C, Melting point of the bullet = 327∘^{\circ}C, Latent heat of fusion of lead = 2.5 ×\times 104 J kg−^{-}1, Specific heat capacity of lead = 125 J/kg K)

Show answer and solution

Answer: (b)

  1. 0.4×12mv2=m(cΔT+L)0.4 \times \tfrac{1}{2}mv^2 = m(c\Delta T + L)
  2. 0.2 v2=125×200+2.5×104=5×1040.2\,v^2 = 125 \times 200 + 2.5 \times 10^4 = 5 \times 10^4
  3. v2=2.5×105⇒v=500 m s−1v^2 = 2.5 \times 10^5 \Rightarrow v = 500\text{ m s}^{-1}
  4. So the answer is (b) 500 m s−1500\text{ m s}^{-1}.
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Q99JEE Main 2022 · 25 JUN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

A copper block of mass 5.0 kg is heated to a temperature of 500∘^{\circ}C and is placed on a large ice block. What is the maximum amount of ice that can melt? [Specific heat of copper: 0.39 J g−^{-}1 ∘C−^{\circ}\text{C}^{-}1 and latent heat of fusion of water: 335 J g−^{-}1]

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Answer: (c)

  1. Copper cools from 500∘C500^{\circ}\text{C} to 0∘C0^{\circ}\text{C}: Q=5000×0.39×500=9.75×105 JQ = 5000 \times 0.39 \times 500 = 9.75 \times 10^5\text{ J}
  2. Ice melted =9.75×105335≈2910 g= \dfrac{9.75 \times 10^5}{335} \approx 2910\text{ g}
  3. So the answer is (c) 2.9 kg2.9\text{ kg}.
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Q100JEE Main 2022 · 25 JUL · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

A block of ice of mass 120 g at temperature 0∘^{\circ}C is put in 300 g of water at 25∘^{\circ} C. The x g of ice melts as the temperature of the water reaches 0∘^{\circ} C. The value of x is _____________. [Use specific heat capacity of water = 4200 Jkg−1K−^{-1}\text{K}^{-}1, Latent heat of ice = 3.5 ×\times 105 Jkg−^{-}1]

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 90

  1. Heat given by water cooling to 0∘C0^{\circ}\text{C} melts the ice: mwc ΔT=xLm_w c\,\Delta T = xL
  2. 0.3×4200×25=31500 J0.3 \times 4200 \times 25 = 31500\ \text{J}
  3. x=315003.5×105=0.09 kgx = \dfrac{31500}{3.5 \times 10^{5}} = 0.09\ \text{kg} (less than 120 g, so valid)
  4. So x=90 gx = 90\ \text{g}.
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Q101JEE Main 2021 · 27 AUG · SHIFT 2Easy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

The height of Victoria falls is 63 m. What is the difference in temperature of water at the top and at the bottom of fall?

[Given 1 cal = 4.2 J and specific heat of water = 1 cal g−^{-}1 ∘C−^{\circ}\text{C}^{-}1]

Show answer and solution

Answer: (a)

  1. mgh=mcΔT⇒ΔT=ghcmgh = mc\Delta T \Rightarrow \Delta T = \dfrac{gh}{c} with c=4200c = 4200 J/kg K.
  2. ΔT=9.8×634200≈0.147∘C\Delta T = \dfrac{9.8 \times 63}{4200} \approx 0.147^\circ\text{C}
  3. So the answer is (a).
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Q102JEE Main 2021 · 25 JUL · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is:

Figure

Show answer and solution

Answer: (b)

  1. Same heating rate and mass: P=mcdTdt⇒c∝1slopeP = mc\dfrac{dT}{dt} \Rightarrow c \propto \dfrac{1}{\text{slope}}.
  2. Slope of A =1203=40∘C/s= \dfrac{120}{3} = 40^{\circ}\text{C/s}; slope of B =906=15∘C/s= \dfrac{90}{6} = 15^{\circ}\text{C/s}.
  3. cAcB=1540=38\dfrac{c_A}{c_B} = \dfrac{15}{40} = \dfrac{3}{8}
  4. So the answer is (b) 38\dfrac{3}{8}.
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Q103JEE Main 2021 · 26 AUG · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

The temperature of equal masses of three different liquids x, y and z are 10∘^{\circ}C, 20∘^{\circ}C and 30∘^{\circ}C respectively. The temperature of mixture when x is mixed with y is 16∘^{\circ}C and that when y is mixed with z is 26∘^{\circ} C. The temperature of mixture when x and z are mixed will be:

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Answer: (c)

  1. Equal masses: x and y: cx(16−10)=cy(20−16)⇒cy=1.5cxc_x(16 - 10) = c_y(20 - 16) \Rightarrow c_y = 1.5c_x.
  2. y and z: cy(26−20)=cz(30−26)⇒cz=1.5cy=2.25cxc_y(26 - 20) = c_z(30 - 26) \Rightarrow c_z = 1.5c_y = 2.25c_x.
  3. x and z: cx(T−10)=2.25cx(30−T)⇒3.25T=77.5c_x(T - 10) = 2.25c_x(30 - T) \Rightarrow 3.25T = 77.5
  4. T≈23.84∘CT \approx 23.84^{\circ}\text{C}
  5. So the answer is (c) 23.84∘C23.84^{\circ}\text{C}.
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Q104NEET 2020 · PHASE 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

The quantities of heat required to raise the temperature of two solid copper spheres of radii r1 and r2 (r1 = 1.5r2) through 1 K are in the ratio:

Show answer and solution

Answer: (d)

  1. Heat for 1 K1\text{ K} =mc= mc, and m=ρ⋅43πr3m = \rho \cdot \tfrac{4}{3}\pi r^3, so Q∝r3Q \propto r^3 (same material).
  2. Q1Q2=(r1r2)3=(1.5)3=278\dfrac{Q_1}{Q_2} = \left(\dfrac{r_1}{r_2}\right)^3 = (1.5)^3 = \dfrac{27}{8}
  3. So the answer is (d) 278\dfrac{27}{8}.
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Q105JEE Main 2020 · 5 SEP · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 cal/(g –oC) (1 cal = 4.2 × 107 ergs) close to:

Show answer and solution

Answer: (a)

  1. KE =12(0.005)(210)2=110.25 J= \tfrac{1}{2}(0.005)(210)^2 = 110.25\text{ J}; heat in bullet =55.125 J= 55.125\text{ J}.
  2. c=0.030 cal/g∘C=0.126 J/g∘Cc = 0.030\text{ cal/g}^{\circ}\text{C} = 0.126\text{ J/g}^{\circ}\text{C}, so mc=5×0.126=0.63 J/∘Cmc = 5 \times 0.126 = 0.63\text{ J/}^{\circ}\text{C}.
  3. ΔT=55.1250.63=87.5∘C\Delta T = \dfrac{55.125}{0.63} = 87.5^{\circ}\text{C}
  4. So the answer is (a) 87.5∘C87.5^{\circ}\text{C}.
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Q106JEE Main 2020 · 4 SEP · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

The specific heat of water

= 4200 J kg-1K-1 and the latent heat of

ice = 3.4 ×\times 105 J kg–1. 100 grams of ice at

0oC is placed in 200 g of water at 25oC. The

amount of ice that will melt as the temperature

of water reaches 0oC is close to (in grams):

Show answer and solution

Answer: (b)

  1. Heat from water cooling to 0∘C0^{\circ}\text{C}: 0.2×4200×25=21000 J0.2 \times 4200 \times 25 = 21000\text{ J}
  2. Ice melted =210003.4×105=0.0618 kg≈61.7 g= \dfrac{21000}{3.4 \times 10^5} = 0.0618\text{ kg} \approx 61.7\text{ g}
  3. So the answer is (b) 61.7.
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Q107JEE Main 2020 · 3 SEP · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

A calorimeter of water equivalent 20 g contains 180 g of water at 25oC. ‘m’ grams of steam at 100oC is mixed in it till the temperature of the mixure is 31oC. The value of ‘m’ is close to: (Latent heat of water = 540 cal g–1, specific heat of water = 1 cal g–1oC–1)

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Answer: (b)

  1. Heat gained by water + calorimeter =(180+20)×1×(31−25)=1200 cal= (180 + 20) \times 1 \times (31 - 25) = 1200\text{ cal}
  2. Heat given by steam =m(540+100−31)=609m= m(540 + 100 - 31) = 609m
  3. 609m=1200⇒m≈1.97≈2 g609m = 1200 \Rightarrow m \approx 1.97 \approx 2\text{ g}
  4. So the answer is (b) 2.
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Q108JEE Main 2020 · 8 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

Three containers C1, C2 and C3 have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in litres) are taken from each containers and mixed (assume no loss of heat during the process)

Figure

The value of θ\theta (in °C to the nearest integer) is ..........

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 50

  1. For mixing equal-specific-heat water: T=∑ViTi∑ViT = \dfrac{\sum V_i T_i}{\sum V_i}.
  2. Row 1: T1+2T2=180T_1 + 2T_2 = 180; Row 2: T2+2T3=90T_2 + 2T_3 = 90; Row 3: 2T1+T3=1802T_1 + T_3 = 180
  3. Adding: 3(T1+T2+T3)=450⇒T1+T2+T3=1503(T_1+T_2+T_3) = 450 \Rightarrow T_1+T_2+T_3 = 150
  4. θ=T1+T2+T33=50∘C\theta = \dfrac{T_1+T_2+T_3}{3} = 50^{\circ}\text{C}
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Q109JEE Main 2020 · 7 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

M grams of steam at 100oC is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produced liquid water at 40oC [heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g] the value of M is ____

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 40

  1. Heat lost by steam (condense + cool to 40∘40^{\circ}C) = heat gained by ice (melt + warm to 40∘40^{\circ}C).
  2. M(540+1×60)=200(80+1×40)M(540 + 1 \times 60) = 200(80 + 1 \times 40)
  3. 600M=24000600M = 24000
  4. So M=40 gM = 40\ \text{g}.
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Q110JEE Main 2019 · 12 APR · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

One kg of water, at 20oC, heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20 Ω\Omega. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to: [Specific heat of water = 4200 J/(kgoC), Latent heat of water = 2260 kJ/kg]

Show answer and solution

Answer: (b)

  1. Power P=V2R=200220=2000 WP = \dfrac{V^2}{R} = \dfrac{200^2}{20} = 2000\text{ W}
  2. Heat =mcΔT+mL=1×4200×80+2.26×106=2.596×106 J= mc\Delta T + mL = 1 \times 4200 \times 80 + 2.26 \times 10^6 = 2.596 \times 10^6\text{ J}
  3. t=2.596×1062000=1298 s≈21.6 mint = \dfrac{2.596 \times 10^6}{2000} = 1298\text{ s} \approx 21.6\text{ min}
  4. So the answer is (b) 22 minutes.
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Q111JEE Main 2019 · 9 APR · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K)

Show answer and solution

Answer: (c)

  1. Spring energy =12kx2=12(800)(0.02)2=0.16 J= \tfrac{1}{2}kx^2 = \tfrac{1}{2}(800)(0.02)^2 = 0.16\text{ J}, all turned into heat.
  2. Heat capacity =1×4184+0.5×400=4384 J/K= 1 \times 4184 + 0.5 \times 400 = 4384\text{ J/K}
  3. ΔT=0.164384≈3.6×10−5 K\Delta T = \dfrac{0.16}{4384} \approx 3.6 \times 10^{-5}\text{ K}
  4. So the answer is (c) 10−5 K10^{-5}\text{ K}.
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Q112JEE Main 2019 · 12 APR · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

When M1 gram of ice at –10oC (specific heat = 0.5 cal g–1oC–1) is added to M2 gram of water at 50°C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g–1 is:

Show answer and solution

Answer: (a)

  1. Heat gained by ice: M1×0.5×10+M1LM_1 \times 0.5 \times 10 + M_1L
  2. Heat lost by water: M2×1×50M_2 \times 1 \times 50
  3. 5M1+M1L=50M2⇒L=50M2M1−55M_1 + M_1L = 50M_2 \Rightarrow L = \dfrac{50M_2}{M_1} - 5
  4. So the answer is (a).
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Q113JEE Main 2019 · 11 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

When 100 g of a liquid A at 100oC is added to 50 g of a liquid B at temperature 75oC, the temperature of the mixture becomes 90oC. The temperature of the mixture, if 100 g of liquid A at 100oC is added to 50 g of liquid B at 50oC, will be:

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Answer: (d)

  1. First mixing: 100 sA(100−90)=50 sB(90−75)⇒sB=43sA100\,s_A(100 - 90) = 50\,s_B(90 - 75) \Rightarrow s_B = \tfrac{4}{3}s_A
  2. Second mixing: 100 sA(100−T)=50⋅43sA(T−50)100\,s_A(100 - T) = 50 \cdot \tfrac{4}{3}s_A(T - 50)
  3. 300(100−T)=200(T−50)⇒T=80∘C300(100 - T) = 200(T - 50) \Rightarrow T = 80^{\circ}\text{C}
  4. So the answer is (d) 80∘C80^{\circ}\text{C}.
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Q114JEE Main 2019 · 11 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

A metal ball of mass 0.1 kg is heated upto 500oC and dropped into a vessel of heat capacity 800 JK–1 and containing 0.5 kg water. The initial temperature of water and vessel is 30oC. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200 Jkg–1 K–1 and 400 Jkg–1 K–1]

Show answer and solution

Answer: (a)

  1. Ball: mc=0.1×400=40 J/Kmc = 0.1 \times 400 = 40\text{ J/K}; water + vessel: 0.5×4200+800=2900 J/K0.5 \times 4200 + 800 = 2900\text{ J/K}.
  2. 40(500−T)=2900(T−30)⇒T≈36.4∘C40(500 - T) = 2900(T - 30) \Rightarrow T \approx 36.4^{\circ}\text{C}
  3. Percentage rise =6.430×100≈21%= \dfrac{6.4}{30} \times 100 \approx 21\%
  4. So the answer is (a) 20%20\%.
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Q115JEE Main 2019 · 11 JAN · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

Ice at –20oC is added to 50 g of water at 40oC. When the temperature of the mixture reaches 0oC, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2J/g/oC Specific heat of Ice = 2.1J/g/oC Heat of fusion of water at 0oC= 334J/g)

Show answer and solution

Answer: (d)

  1. Heat lost by water =50×4.2×40=8400 J= 50 \times 4.2 \times 40 = 8400\text{ J}
  2. Ice of mass mm: warming m×2.1×20=42mm \times 2.1 \times 20 = 42m; melting (m−20)×334(m - 20) \times 334.
  3. 42m+334(m−20)=8400⇒376m=15080⇒m≈40 g42m + 334(m - 20) = 8400 \Rightarrow 376m = 15080 \Rightarrow m \approx 40\text{ g}
  4. So the answer is (d) 40 g.
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Q116JEE Main 2019 · 10 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

An unknown metal of mass 192 g heated to a temperature of 100oC was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4oC. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5oC. (Specific heat of brass is 394 J kg–1 K–1)

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Answer: (d)

  1. Heat lost by metal = heat gained by water + brass calorimeter.
  2. 0.192 s (100−21.5)=(0.128×394+0.240×4200)(21.5−8.4)0.192\,s\,(100-21.5) = (0.128 \times 394 + 0.240 \times 4200)(21.5-8.4)
  3. 15.072 s=1058.4×13.1≈1386515.072\,s = 1058.4 \times 13.1 \approx 13865
  4. s≈920 J kg−1K−1s \approx 920\ \text{J kg}^{-1}\text{K}^{-1}, closest option.
  5. So the answer is (d) 916 J kg−1K−1916\ \text{J kg}^{-1}\text{K}^{-1}.
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Q117JEE Main 2019 · 8 APR · SHIFT 1Difficult

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

A thermally insulated vessel contains 150g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to: (Latent heat of vaporization of water = 2.10 × 106 J kg–1 and Latent heat of Fusion of water = 3.36 × 105 J kg–1)

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Answer: (c)

  1. Heat for evaporation comes from water freezing: mvLv=miLfm_v L_v = m_i L_f, with mv+mi=150 gm_v + m_i = 150\ \text{g}.
  2. mv=150×LfLf+Lv=150×3.36×1053.36×105+2.10×106m_v = \dfrac{150 \times L_f}{L_f + L_v} = \dfrac{150 \times 3.36 \times 10^{5}}{3.36 \times 10^{5} + 2.10 \times 10^{6}}
  3. mv=150×0.3362.436≈20.7 gm_v = \dfrac{150 \times 0.336}{2.436} \approx 20.7\ \text{g}
  4. So the answer is (c) 20 g.
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Q118JEE Main 2017 · 8 APR · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

In an experiment, a sphere of aluminium of mass 0.20 kg is heated upto 150oC. Immediately, it is put into water of volume 150 cc at 27oC kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40oC. The specific heat of aluminium is: (take 4.2 Joule = 1 calorie)

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Answer: (d)

  1. Water mass =150 g=0.150 kg= 150\ \text{g} = 0.150\ \text{kg}; add water equivalent 0.025 kg0.025\ \text{kg}.
  2. Heat lost by Al = heat gained: 0.20 s (150−40)=(0.150+0.025)(4200)(40−27)0.20\,s\,(150-40) = (0.150+0.025)(4200)(40-27)
  3. 22 s=0.175×4200×13=955522\,s = 0.175 \times 4200 \times 13 = 9555
  4. s≈434 J/kg∘Cs \approx 434\ \text{J/kg}^{\circ}\text{C}
  5. So the answer is (d) 434 J/kg∘C434\ \text{J/kg}^{\circ}\text{C}.
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Q119JEE Main 2017Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75oC. T is given by: (Given: room temperature = 30oC, specific heat of copper = 0.1 cal/gmoC)

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Answer: (c)

  1. Heat lost by ball = heat gained by calorimeter + water.
  2. 100(0.1)(T−75)=[100(0.1)+170(1)](75−30)100(0.1)(T-75) = [100(0.1) + 170(1)](75-30)
  3. 10(T−75)=180×45=8100⇒T−75=81010(T-75) = 180 \times 45 = 8100 \Rightarrow T - 75 = 810
  4. So the answer is (c) T=885∘CT = 885^{\circ}\text{C}.
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Q120JEE Main 2016 · 9 APR · SHIFT 1Easy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

200 g water is heated from 40oC to 60oC. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kg/K):

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Answer: (c)

  1. No work done, so ΔU=Q=mcΔT\Delta U = Q = mc\Delta T.
  2. ΔU=0.2×4184×20=16736\Delta U = 0.2 \times 4184 \times 20 = 16736 J
  3. So the answer is (c) 16.7 kJ.
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Q121JEE Main 2015 · 11 APR · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

An experiment takes 10 minutes to raise the temperature of water in a container from 0oC to 100oC and another 55 minutes to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of water to be 1 cal/goC, the heat of vapourization according to this experiment will come out to be:

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Answer: (c)

  1. Heat rate =m×1×10010=10m cal/min= \dfrac{m \times 1 \times 100}{10} = 10m\text{ cal/min}
  2. In 55 min: mL=55×10m⇒L=550 cal/gmL = 55 \times 10m \Rightarrow L = 550\text{ cal/g}
  3. So the answer is (c) 550 cal/g550\text{ cal/g}.
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Q122JEE Main 2014 · 9 APR · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160J/s. In how much time will the temperature of water rise from 27oC to 77oC? (Specific heat of water is 4.2 kJ/kg K and that of the container is negligible).

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Answer: (a)

  1. Heat needed =2×4200×50=4.2×105 J= 2 \times 4200 \times 50 = 4.2 \times 10^5\text{ J}
  2. Net power =1000−160=840 W= 1000 - 160 = 840\text{ W}
  3. t=4.2×105840=500 s=8 min 20 st = \dfrac{4.2 \times 10^5}{840} = 500\text{ s} = 8\text{ min } 20\text{ s}
  4. So the answer is (a).
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Q123JEE Main 2013 · 22 APR · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Latent heat, phase change & heating curve

Given that 1 g of water in liquid phase has volume 1 cm3^{3} and in vapour phase 1671 cm3^{3} at atmospheric pressure and the latent heat of vaporization of water is 2256 J/g; the change in the internal energy in joules for 1 g of water at 373 K when it changes from liquid phase to vapour phase at the same temperature is:

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Answer: (c)

  1. ΔU=Q−PΔV\Delta U = Q - P\Delta V, with Q=2256 JQ = 2256\text{ J}.
  2. PΔV=1.013×105×1670×10−6≈167 JP\Delta V = 1.013 \times 10^5 \times 1670 \times 10^{-6} \approx 167\text{ J}
  3. ΔU=2256−167=2089 J\Delta U = 2256 - 167 = 2089\text{ J}
  4. So the answer is (c) 2089.
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Q124JEE Main 2013 · 23 APR · SHIFT 1Difficult

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

500 g of water and 100 g of ice at 0°C are in a calorimeter whose water equivalent is 40 g. 10 g of steam at 100°C is added to it. Then water in the calorimeter is: (Latent heat of ice = 80 cal/g, Latent heat of steam = 540 cal/g)

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Answer: (b)

  1. Heat steam can give down to 0∘0^{\circ}C: 10(540+100)=6400 cal10(540 + 100) = 6400\ \text{cal}.
  2. Heat to melt all ice: 100×80=8000 cal>6400100 \times 80 = 8000\ \text{cal} > 6400, so final temperature is 0∘0^{\circ}C.
  3. Ice melted =640080=80 g= \dfrac{6400}{80} = 80\ \text{g}
  4. Water =500+80+10=590 g= 500 + 80 + 10 = 590\ \text{g}
  5. So the answer is (b) 590 g.
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Q125JEE Main 2011 · AIEEEEasy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

100 g100\text{ g} of water is heated from 30∘C30^{\circ}\text{C} to 50∘C50^{\circ}\text{C}. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K4184\text{ J/kg/K}):

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Answer: (a)

  1. No work done, so ΔU=Q=mcΔT\Delta U = Q = mc\Delta T.
  2. ΔU=0.1×4184×20=8368\Delta U = 0.1 \times 4184 \times 20 = 8368 J
  3. So the answer is (a) 8.4 kJ.
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Q126JEE Main 2004 · AIEEEEasy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

Time taken by a 836 W836\text{ W} heater to heat one litre of water from 10∘C10^{\circ}\text{C} to 40∘C40^{\circ}\text{C} is

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Answer: (a)

  1. Heat needed Q=mcΔTQ = mc\Delta T with m=1 kgm = 1\text{ kg} (1 litre of water).
  2. Q=1×4180×30=125400 JQ = 1 \times 4180 \times 30 = 125400\text{ J}
  3. t=QP=125400836=150 st = \dfrac{Q}{P} = \dfrac{125400}{836} = 150\text{ s}
  4. So the answer is (a) 150 s150\text{ s}.
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Q127JEE Main 2002 · AIEEEEasy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

Heat given to a body which raises its temperature by 1∘C1^{\circ}\text{C} is

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Answer: (b)

  1. Thermal (heat) capacity =mc= mc is the heat needed to raise the temperature of the whole body by 1∘C1^{\circ}\text{C}.
  2. Specific heat is per unit mass; water equivalent is a mass; temperature gradient is dT/dxdT/dx.
  3. So the answer is (b) thermal capacity.
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Q128NEET 1999 · AIPMTMedium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

If 1 g1\text{ g} of steam is mixed with 1 g1\text{ g} of ice, then the resultant temperature of the mixture is

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Answer: (c)

  1. Ice needs 80+100=180 cal80 + 100 = 180\text{ cal} to become water at 100∘C100^{\circ}\text{C}.
  2. Steam gives 540 cal540\text{ cal} just by condensing at 100∘C100^{\circ}\text{C}, which is more than 180 cal180\text{ cal}.
  3. So only part of the steam condenses and the mixture stays at 100∘C100^{\circ}\text{C}.
  4. So the answer is (c) 100∘C100^{\circ}\text{C}.
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Q129NEET 1990 · AIPMTEasy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Specific heat, thermal capacity & Joule's law

Thermal capacity of 40 g of aluminium (s=0.2 cal/g Ks = 0.2\text{ cal/g K}) is

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Answer: (d)

  1. C=ms=40×0.2=8C = ms = 40 \times 0.2 = 8 cal/K
  2. 8×4.2=33.68 \times 4.2 = 33.6 J/K
  3. So the answer is (d).
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Q130NEET 1988 · AIPMTMedium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat · Principle of calorimetry: mixtures

10 gm of ice cubes at 0∘C0^{\circ}\text{C} are released in a tumbler (water equivalent 55 g) at 40∘C40^{\circ}\text{C}. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L=80 cal/gL = 80\text{ cal/g})

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Answer: (b)

  1. Heat lost by tumbler = heat to melt ice + warm melt water.
  2. 55(40−T)=10×80+10T55(40 - T) = 10 \times 80 + 10T
  3. 2200−55T=800+10T⇒T=140065≈21.5∘C2200 - 55T = 800 + 10T \Rightarrow T = \dfrac{1400}{65} \approx 21.5^{\circ}\text{C}
  4. So the answer is (b) 22∘C22^{\circ}\text{C}.
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