JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Heat & Thermodynamics

A Carnot engine whose heat sinks at

2 7^{\circ} C

, has an efficiency of

25%

. By how many degrees should the temperature of the source be changed to increase the efficiency by

100%

of the original efficiency ?

Answer: (B)

Solution

For a Carnot engine, efficiency

η = 1 - \frac{T _{C}}{T _{H}}

where temperatures are in Kelvin.

Initially:

T_{C} = 300

η = 0.25

0.25 = 1 - \frac{300}{T _{H}} \Rightarrow \frac{300}{T _{H}} = 0.75 \Rightarrow T_{H} = 400

K

Increasing efficiency by 100% of original means new efficiency =

0.25 + 0.25 = 0.50

0.50 = 1 - \frac{300}{T _{H}^{'}} \Rightarrow \frac{300}{T _{H}^{'}} = 0.50 \Rightarrow T_{H}^{'} = 600

K

Increase in temperature =

600 - 400 = 200

K =

20 0^{\circ}

C

The answer is (B).

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option