JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Heat & Thermodynamics

A Carnot engine with efficiency

50%

takes heat from a source at 600 K. In order to increase the efficiency to

70%

, keeping the temperature of sink same, the new temperature of the source will be:

(A)360 K
(B)1000 K✓ Correct
(C)900 K
(D)300 K

Answer: (B)

Solution

A Carnot engine has efficiency

η = 1 - \frac{T _{2}}{T _{1}}

. The first condition

\frac{1}{2} = 1 - \frac{T _{2}}{600}

fixes the sink at

T_{2} = 300 K

. Keeping this sink and demanding

η = 0.7

gives

\frac{7}{10} = 1 - \frac{300}{T _{1}^{'}}

, so

\frac{300}{T _{1}^{'}} = \frac{3}{10}

. The source must be raised to

T_{1}^{'} = 1000 K

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Single correct option

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