JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Heat & Thermodynamics

WhatsApp Email

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

1 0^{\circ} C

and

2 0^{\circ} 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:

(A)A, B, C only✓ Correct
(B)A, B only
(C)A, C only
(D)B, C, D only

Answer: (A)

Solution

Statement A: By Newton's law of cooling,

\frac{d θ}{d t} \propto Δ T

, so doubling

Δ T

doubles the rate — TRUE. Statement B: Using Stefan-Boltzmann law,

\frac{H _{P}}{H _{Q}} = (\frac{283}{293})^{4} \approx 1 : 1.15

— TRUE. Statement C: Carnot efficiency

η = 1 - \frac{T _{L}}{T _{H}} = 1 - \frac{100}{400} = 0.75 = 75%

— TRUE. Statement D: Quadrupling

Δ T

would quadruple the rate, not double it — FALSE. Thus A, B, C are correct.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Heat & Thermodynamics →

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Heat & Thermodynamics

Solution

More JEE Main questions on Heat & Thermodynamics