Read the following statements :
A. When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice.
B. Two bodies P and Q having equal surface areas are maintained at temperature 10° C and 20° C. The thermal radiation emitted in a given time by P and Q are in the ratio 1: 1.15.
C. A Carnot Engine working between 100 K and 400 K has an efficiency of 75 %.
D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice.
Choose the correct answer from the options given below :

Solution

Physics sir · 3 steps

Test each statement with its own law, and switch to kelvin before raising to the fourth power.

1. Newton's law. For a small temperature difference, the rate of heat loss is proportional to Δ T. Doubling Δ T doubles the rate, so A is true. Quadrupling Δ T makes the rate four times, not twice, so D is false.
2. Stefan's law. Radiated energy goes as T⁴ with T in kelvin. P is at 283 K and Q is at 293 K, and both have the same area and time.

   E_Q/E_P = (293/283)⁴ = (1.0353)⁴ ≈ 1.15

   Statement	Check	Verdict
   A	rate ∝ Δ T, so 2Δ T gives 2× rate	true
   B	E_P : E_Q = 1 : 1.15	true
   C	η = 1 - 100/400 = 0.75 = 75%	true
   D	4Δ T gives 4× rate, not 2×	false

3. So. A, B and C are true and D is false, which is option (a).

The idea to keep. Newton's cooling is linear in the temperature difference. Stefan's law and Carnot efficiency both need absolute temperature, so convert Celsius to kelvin first.

Answer from the NTA final answer key. Solution by PrepWiser.