A heat engine operates between a cold reservoir at temperature $T_{2}=400 \mathrm{~K}$ and a hot reservoir at temperature $\mathrm{T}_{1}$. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ____ K.

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Heat & Thermodynamics

A heat engine operates between a cold reservoir at temperature

T_{2} = 400 K

and a hot reservoir at temperature

T_{1}

. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ____ K.

Answer: 500

Solution

Work done by the engine:

W = Q_{1} - Q_{2} = 300 - 240 = 60

J. Efficiency:

η = \frac{W}{Q _{1}} = \frac{60}{300} = 0.2

. For a Carnot engine (maximum efficiency),

η = 1 - \frac{T _{2}}{T _{1}}

. Thus

0.2 = 1 - \frac{400}{T _{1}}

, giving

\frac{400}{T _{1}} = 0.8

, so

T_{1} = \frac{400}{0.8} = 500

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Integer answer

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