Heat Transfer
ICSE · Class 11 · Physics
Most important questions from Heat Transfer for ICSE Class 11 Physics board exam 2026. MCQs, short answer, and long answer questions with marks.
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In Searle's method to find thermal conductivity, four thermometers T₁, T₂, T₃, T₄ are used. Which thermometers measure the temperature of the rod, and which measure the temperature of flowing water?
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T₁, T₂ measure rod temperature; T₃, T₄ measure water temperature
Step 1: In Searle's apparatus, thermometers T₁ and T₂ are inserted into holes C and D drilled in the rod itself (with mercury for good thermal contact). They record the steady-state temperature gradient along the rod. Step 2: Thermometers T₃ and T₄ measure the inlet and outlet temperatures of the cooling water flowing through the spiral copper tube wound at the cold end. Step 3: The formula K = mc(θ₄ - θ₃)l / [A(θ₁ - θ₂)t] uses all four readings to calculate K. The wrong options confuse the arrangement of the thermometers.
A black body at 727°C emits heat at a certain rate. If its temperature is reduced to 227°C, the new rate of heat emission will be what fraction of the original rate?
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1/16
Step 1: Convert temperatures to Kelvin: T₁ = 727 + 273 = 1000 K, T₂ = 227 + 273 = 500 K. Step 2: By Stefan's law, E ∝ T⁴. So E₂/E₁ = (T₂/T₁)⁴ = (500/1000)⁴ = (1/2)⁴ = 1/16. Step 3: The new emission rate is 1/16 of the original. Common error: Using Celsius instead of Kelvin gives (227/727)⁴ ≈ 1/136, which is wrong. Always convert to Kelvin for Stefan's law.
Two slabs of equal length l and same cross-sectional area A are connected in parallel. Their thermal conductivities are K₁ and K₂. What is the equivalent thermal conductivity?
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(K₁ + K₂) / 2
Step 1: For parallel combination with equal areas A₁ = A₂ = A, the equivalent conductivity formula is: K = (K₁A₁ + K₂A₂) / (A₁ + A₂). Step 2: With A₁ = A₂ = A: K = (K₁A + K₂A) / (2A) = (K₁ + K₂) / 2. This is the arithmetic mean of the two conductivities. Wrong options: K₁K₂/(K₁+K₂) is the formula for electrical resistors in parallel; 2K₁K₂/(K₁+K₂) is the formula for series combination of equal-thickness slabs; (K₁+K₂) is not a valid formula.
A hot body at temperature T is placed in a room at temperature T₀. According to Newton's law of cooling, the rate of fall of temperature of the body is proportional to:
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T - T₀
Step 1: Newton's law of cooling states that the rate of loss of heat is proportional to the temperature difference between the body and its surroundings: -dQ/dt ∝ (T - T₀). Step 2: Since dQ = ms dT, this gives: -dT/dt = K(T - T₀), where K is a positive constant. Step 3: So the rate of fall of temperature is proportional to (T - T₀), which is the simple temperature difference. Note: The T⁴ - T₀⁴ term appears in Stefan-Boltzmann law for net radiation, not in Newton's law of cooling. Newton's law is valid only for small temperature differences.
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