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A carrot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is:

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200°C

15°C

27°C

100°C

answer is B.

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Detailed Solution

$\begin{array}{l} η = 1 - \frac{T_{2}}{T_{1}} \\ 0.5 = 1 - \frac{T_{2}}{327 + 273} \\ 0.5 = 1 - \frac{T_{2}}{600} \\ \frac{T_{2}}{600} = \frac{1}{2} \\ T_{2} = 300 K \\ = 300 - 273 \\ = 27^{\circ} C \end{array}$

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