For an ideal heat engine, the temperature of the source is $127^{\circ} \mathrm{C}$. In order to have $60 \%$ efficiency the temperature of the sink should be ⋯⋯----- ${ }^{\circ} \mathrm{C}$. (Round off to the Nearest Integer)

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Heat & Thermodynamics

For an ideal heat engine, the temperature of the source is

12 7^{\circ} C

. In order to have

60%

efficiency the temperature of the sink should be ⋯⋯-----

^{\circ} C

. (Round off to the Nearest Integer)

Answer: -113

Solution

An ideal engine working between a source at

T_{1}

and a sink at

T_{2}

has efficiency

η = 1 - \frac{T _{2}}{T _{1}}

with both temperatures on the absolute scale. Here

T_{1} = 12 7^{\circ} C = 400 K

and

η = 0.60

, so

\frac{T _{2}}{400} = 1 - 0.60 = 0.40 \Rightarrow T_{2} = 160 K

On the Celsius scale

160 - 273 = - 113

, so the sink must be held at

- 113^{\circ} C

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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