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Heat, temperature, expansion, specific heat, calorimetry and latent heat: NEET MCQs with answers

75 questions on this topic of Properties of Solids and Liquids (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Taking the specific heat of water as 4.2 J g⁻¹ °C⁻¹, the heat needed to raise the temperature of 300 g of water by 40 °C is:

  1. 50.4 J
  2. 1260 J
  3. 12000 J
  4. 50400 J
Show answer and explanation

Answer: (D) 50400 J

Heat absorbed without a change of state is Q = mcΔT = 300 × 4.2 × 40 = 50400 J.

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2. A pendulum clock keeps correct time at 20 °C. The linear expansion coefficient of the pendulum's metal rod is 2 × 10⁻⁵ °C⁻¹. How much time will the clock lose per day if the temperature increases to 40 °C?

  1. 8.64 s
  2. 17.28 s
  3. 34.56 s
  4. 4.32 s
Show answer and explanation

Answer: (B) 17.28 s

Fractional change in period is ΔT/T = (1/2)αΔθ = (1/2)(2 × 10⁻⁵)(20) = 2 × 10⁻⁴. The time lost per day is Δt = (ΔT/T) × 86400 s = 2 × 10⁻⁴ × 86400 = 17.28 s.

hard

3. A piece of ice at 0 °C falls from a height h such that upon hitting the ground, all of its mechanical energy is converted into heat within the ice. What is the minimum height from which it must fall so that it melts completely? (Latent heat of fusion of ice L = 3.36 × 10⁵ J kg⁻¹, g = 9.8 m s⁻²)

  1. 34.3 km
  2. 3.43 km
  3. 17.1 km
  4. 68.6 km
Show answer and explanation

Answer: (A) 34.3 km

The potential energy lost by mass m is m·g·h. This energy is converted into heat to melt the ice, so m·g·h = m·L. Therefore, h = L / g = (3.36 × 10⁵ J kg⁻¹) / (9.8 m s⁻²) ≈ 34286 m ≈ 34.3 km.

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4. A block of ice of mass 100 g at 0 °C is dropped into a copper calorimeter containing 100 g of water at 20 °C. The water equivalent of the calorimeter is 10 g. Assuming no heat is lost to the surroundings, how much ice remains unmelted at thermal equilibrium? (Latent heat of fusion of ice = 80 cal g⁻¹, specific heat of water = 1 cal g⁻¹ °C⁻¹)

  1. 72.5 g
  2. 27.5 g
  3. 50.0 g
  4. 80.0 g
Show answer and explanation

Answer: (A) 72.5 g

Total heat lost by water and calorimeter in cooling from 20 °C to 0 °C = (100 + 10) × 1 × (20 - 0) = 2200 cal. Mass of ice melted m = Q / L = 2200 / 80 = 27.5 g. Therefore, mass of unmelted ice remaining = 100 - 27.5 = 72.5 g.

easy

5. Two identical containers contain equal masses of two different liquids, A and B. Both are supplied heat at the same constant rate. The graph of temperature versus time shows that the slope for liquid A is twice the slope for liquid B. Neglecting heat capacities of the containers, the ratio of the specific heat capacity of A to that of B (s_A / s_B) is:

  1. 2
  2. 1/2
  3. 1/4
  4. 4
Show answer and explanation

Answer: (B) 1/2

Rate of heat supplied is dQ/dt = P = m s (dT/dt). Since mass m and power P are identical, s × (dT/dt) = constant. Therefore, s_A / s_B = (dT/dt)_B / (dT/dt)_A = 1/2.

hard

6. A circular ring of wire of radius R and cross-sectional area A has a coefficient of linear expansion α and Young's modulus Y. If the temperature of the ring is raised by ΔT, what is the tension developed in the ring if it is prevented from expanding radially by a rigid, non-conducting circular core?

  1. Y A α ΔT / 2
  2. 2π Y A α ΔT
  3. Y A α ΔT
  4. Zero
Show answer and explanation

Answer: (C) Y A α ΔT

The unrestrained expansion in circumference would be ΔC = 2π R α ΔT. Since it is prevented from expanding, the compressive circumferential strain is ΔC/C = (2π R α ΔT)/(2π R) = α ΔT. The restoring compressive/tensile stress developed along the ring circumference is σ = Y × strain = Y α ΔT. Therefore, the force or tension developed along the cross-section is T_force = σ A = Y A α ΔT.

easy

7. Taking the specific heat of water as 4.2 J g⁻¹ °C⁻¹, the heat needed to raise the temperature of 350 g of water by 50 °C is:

  1. 1470 J
  2. 73.5 J
  3. 73500 J
  4. 17500 J
Show answer and explanation

Answer: (C) 73500 J

Heat absorbed without a change of state is Q = mcΔT = 350 × 4.2 × 50 = 73500 J.

easy

8. Taking the specific heat of water as 4.2 J g⁻¹ °C⁻¹, the heat needed to raise the temperature of 350 g of water by 40 °C is:

  1. 58.8 J
  2. 14000 J
  3. 1470 J
  4. 58800 J
Show answer and explanation

Answer: (D) 58800 J

Heat absorbed without a change of state is Q = mcΔT = 350 × 4.2 × 40 = 58800 J.

easy

9. Taking the specific heat of water as 4.2 J g⁻¹ °C⁻¹, the heat needed to raise the temperature of 400 g of water by 35 °C is:

  1. 58.8 J
  2. 1680 J
  3. 14000 J
  4. 58800 J
Show answer and explanation

Answer: (D) 58800 J

Heat absorbed without a change of state is Q = mcΔT = 400 × 4.2 × 35 = 58800 J.

easy

10. A platinum resistance thermometer has a resistance of 5.00 Ω at 0 °C and 5.93 Ω at 100 °C. When the thermometer is inserted into a hot bath, its resistance is measured to be 5.70 Ω. Assuming the resistance varies linearly with temperature, the temperature of the bath is:

  1. 81.2 °C
  2. 68.5 °C
  3. 75.3 °C
  4. 70.0 °C
Show answer and explanation

Answer: (C) 75.3 °C

The temperature t on the resistance scale is given by t = [(R_t - R₀) / (R₁₀₀ - R₀)] × 100 °C. Substituting the given values: t = [(5.70 - 5.00) / (5.93 - 5.00)] × 100 = (0.70 / 0.93) × 100 ≈ 75.27 °C ≈ 75.3 °C.

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11. A composite rod is made by joining two rods of different materials end to end. The rods have lengths L₁ and L₂ and linear thermal expansion coefficients α₁ and α₂ respectively. The effective coefficient of linear expansion of the composite rod is:

  1. (L₁α₂ + L₂α₁) / (L₁ + L₂)
  2. (α₁ + α₂) / 2
  3. √(α₁α₂)
  4. (L₁α₁ + L₂α₂) / (L₁ + L₂)
Show answer and explanation

Answer: (D) (L₁α₁ + L₂α₂) / (L₁ + L₂)

Total initial length is L = L₁ + L₂. Change in total length is ΔL = ΔL₁ + ΔL₂ = (L₁α₁ + L₂α₂)ΔT. The effective coefficient of linear expansion α_eff = ΔL / (L ΔT) = (L₁α₁ + L₂α₂) / (L₁ + L₂).

easy

12. If the temperature of a uniform solid sphere is increased by ΔT, the fractional change in its moment of inertia about its diameter is: (Take α as the coefficient of linear expansion of the material)

  1. 2α ΔT
  2. α ΔT
  3. 3α ΔT
  4. 5α ΔT
Show answer and explanation

Answer: (A) 2α ΔT

The moment of inertia of a solid sphere is I = (2/5) M R². Differentiating with respect to temperature while mass remains constant yields ΔI/I = 2(ΔR/R). Since ΔR/R = α ΔT, we have ΔI/I = 2α ΔT.

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