For an ideal heat engine, the temperature of the source is 127 ° C. In order to have 60% efficiency the temperature of the sink should be ___________ ° C. (Round off to the Nearest Integer) Numerical answer type. Work it out, then open the answer.

Answer: -113. T₁ = 127 + 273 = 400 K 0.6 = 1 - (T₂)/(400) ⇒ T₂ = 0.4 × 400 = 160 K T₂ = 160 - 273 = -113° C

For an ideal heat engine, the temperature of the source is 127 ° C… | JEE Main 2021 (16 Mar, Shift 2)

Q42JEE Main 2021 · 16 MAR · SHIFT 2Easy

Heat Engines, Refrigerators & Second Law

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

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: -113

$T_1 = 127 + 273 = 400$ K
$0.6 = 1 - \frac{T_2}{400} \Rightarrow T_2 = 0.4 \times 400 = 160$ K
$T_2 = 160 - 273 = -113^\circ$ C

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