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Heat transfer by conduction, convection and radiation: NEET MCQs with answers

50 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. The SI unit of thermal conductivity (K) is:

  1. W m⁻¹ K⁻¹
  2. W m K⁻¹
  3. J s⁻¹ m⁻² K
  4. W m⁻² K⁻¹
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Answer: (A) W m⁻¹ K⁻¹

From Fourier's law of heat conduction, rate of heat transfer is H = K A (ΔT/L). Rearranging gives K = (H × L) / (A × ΔT). Substituting the SI units gives (W × m) / (m² × K) = W m⁻¹ K⁻¹.

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2. Consider the following two statements regarding thermal radiation: Statement I: A perfectly black body is one which absorbs completely all radiations of all wavelengths falling upon it. Statement II: The absorptive power of a perfectly black body is unity, and its emissivity is zero. In the light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because a perfectly black body absorbs all incident radiation irrespective of wavelength. Statement II is incorrect because for a perfectly black body, both absorptive power (a) and emissivity (e) are equal to 1, not 0 (by Kirchhoff's law of radiation).

hard

3. A metal sphere of radius R, density ρ and specific heat capacity c is heated to a temperature T and suspended in an evacuated enclosure at absolute zero temperature (0 K). If emissivity of the sphere is e, the initial rate of fall of temperature (-dT/dt) is proportional to:

  1. R / T⁴
  2. R² / T⁴
  3. T³ / R²
  4. T⁴ / R
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Answer: (D) T⁴ / R

The rate of radiant heat loss is dQ/dt = eσ(4πR²)T⁴. Also, dQ/dt = mc(-dT/dt) = [(4/3)πR³ρc](-dT/dt). Equating the two gives -dT/dt = [3eσ/(ρc)] × (T⁴/R). Therefore, the initial cooling rate is proportional to T⁴ / R.

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4. Three identical conducting rods of the same material and dimensions are connected in series. The free end of the first rod is maintained at 90 °C and the free end of the third rod is maintained at 0 °C. Assuming steady state and no heat loss from the sides, the temperatures of the two junctions between the rods are:

  1. 70 °C and 35 °C
  2. 45 °C and 22.5 °C
  3. 60 °C and 30 °C
  4. 50 °C and 25 °C
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Answer: (C) 60 °C and 30 °C

Since the rods are identical and connected in series, their thermal resistances are equal (R₁ = R₂ = R₃ = R). The total temperature drop of 90 °C is divided equally across the three identical resistances, giving a drop of 30 °C across each rod. Thus, the junction temperatures are 90 - 30 = 60 °C and 60 - 30 = 30 °C.

easy

5. Given below are two statements: Statement I: A body with high emissivity e is a good absorber of thermal radiation. Statement II: According to Kirchhoff's law of radiation, for any body in thermal equilibrium, spectral emissivity equals spectral absorptivity. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Kirchhoff's law states that at thermal equilibrium, good emitters are good absorbers (e_λ = a_λ). Hence, high emissivity implies high absorptivity, making both statements fully correct.

hard

6. A black body at absolute temperature T is enclosed inside a chamber maintained at temperature T₀ (where T > T₀). If the temperature difference ΔT = T - T₀ is very small compared to T₀, the net rate of heat loss from the black body is proportional to:

  1. T₀⁴ ΔT
  2. T₀² ΔT
  3. T₀³ ΔT
  4. (ΔT)⁴
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Answer: (C) T₀³ ΔT

The net rate of heat loss is dQ/dt = σA(T⁴ - T₀⁴). When T = T₀ + ΔT with ΔT ≪ T₀, T⁴ = (T₀ + ΔT)⁴ ≈ T₀⁴[1 + 4(ΔT/T₀)]. Therefore, T⁴ - T₀⁴ ≈ 4T₀³ΔT. Thus, the net heat loss rate is directly proportional to T₀³ ΔT, which is the foundation of Newton's law of cooling derived from Stefan's law.

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7. A beaker of hot water in an ambient environment at 20 °C cools according to Newton's law of cooling. It takes 6 minutes to cool from 70 °C to 50 °C. The time taken to cool further from 50 °C to 40 °C will be:

  1. 4.0 minutes
  2. 6.0 minutes
  3. 5.0 minutes
  4. 7.5 minutes
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Answer: (C) 5.0 minutes

By Newton's law approximation, (T₁ - T₂)/t = K[(T₁ + T₂)/2 - T₀]. For the first drop: (70 - 50)/6 = K[(60) - 20] => 20/6 = 40K => K = 1/12 min⁻¹. For the second drop: (50 - 40)/t₂ = (1/12)[(45) - 20] => 10/t₂ = 25/12 => t₂ = 120/25 = 4.8 minutes ≈ 5.0 minutes.

easy

8. Sea breeze during the day is an example of heat transfer by:

  1. Convection
  2. Conduction
  3. Radiation only
  4. Evaporation only
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Answer: (A) Convection

Land heats faster than the sea; warm air over land rises and cooler air from the sea moves in. This bulk movement of fluid is natural convection.

NCERT Class 11

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9. Two spherical black bodies A and B radiate energy such that the ratio of their absolute temperatures is T_A : T_B = 2 : 1 and the ratio of their surface areas is A_A : A_B = 1 : 8. The ratio of the total power radiated by body A to that by body B is:

  1. 2 : 1
  2. 1 : 1
  3. 4 : 1
  4. 1 : 2
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Answer: (A) 2 : 1

From the Stefan-Boltzmann law, total radiated power is P = σ A T⁴. Therefore, P_A / P_B = (A_A / A_B) × (T_A / T_B)⁴ = (1/8) × (2)⁴ = (1/8) × 16 = 2. Hence, the ratio is 2 : 1.

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10. A composite slab is made of two layers in contact: layer 1 of thickness 10 cm and thermal conductivity K, and layer 2 of thickness 30 cm and thermal conductivity 3K. If the outer face of layer 1 is at 100 °C and the outer face of layer 2 is at 0 °C, what is the temperature at the interface between the two layers in steady state?

  1. 25 °C
  2. 33.3 °C
  3. 66.7 °C
  4. 50 °C
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Answer: (D) 50 °C

The thermal resistance of layer 1 is R₁ = L₁/(K₁A) = 10/(KA). The thermal resistance of layer 2 is R₂ = L₂/(K₂A) = 30/(3KA) = 10/(KA). Since R₁ = R₂, the temperature drop across each layer is equal: ΔT₁ = ΔT₂ = (100 - 0)/2 = 50 °C. Thus, the interface temperature is 100 - 50 = 50 °C.

easy

11. A metal rod of length 0.5 m has a temperature difference of 100 °C maintained across its two ends in steady state. If the temperature gradient along the rod is uniform, the magnitude of the temperature gradient is:

  1. 50 °C m⁻¹
  2. 100 °C m⁻¹
  3. 200 °C m⁻¹
  4. 400 °C m⁻¹
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Answer: (C) 200 °C m⁻¹

Temperature gradient is defined as the change in temperature per unit length along the heat flow path. Here, temperature gradient = ΔT / L = 100 °C / 0.5 m = 200 °C m⁻¹.

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12. Three identical rods of the same material and cross-sectional area are joined to form a Y-shaped junction. The free ends of the two branches are maintained at 90 °C and 60 °C, while the free end of the stem is maintained at 30 °C. The length of each branch and stem is identical. What is the temperature of the central junction in steady state?

  1. 50 °C
  2. 70 °C
  3. 65 °C
  4. 60 °C
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Answer: (D) 60 °C

Let T_j be the junction temperature. By junction rule for heat flow, the sum of incoming heat currents equals the outgoing heat current: (90 - T_j)(KA/L) + (60 - T_j)(KA/L) = (T_j - 30)(KA/L). Canceling (KA/L): 90 - T_j + 60 - T_j = T_j - 30 ⇒ 150 - 2T_j = T_j - 30 ⇒ 3T_j = 180 ⇒ T_j = 60 °C.

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