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A Carnot engine has an efficiency of 50% when its source is at a temperature $327^\circ C$. The temperature of the sink is:
A
$200^\circ C$
B
$27^\circ C$
C
$15^\circ C$
D
$100^\circ C$
Explanation
$\eta=1-\frac{T_{sink}}{T_{source}}$ gives $T_{sink}=300\ K=27^\circ C$.
Detailed Solution
$\eta=1-\frac{T_{sink}}{T_{source}}=0.5\Rightarrow\frac{T_{sink}}{T_{source}}=0.5$
$\Rightarrow T_{sink}=\frac{1}{2}\times(273+327)=\frac{1}{2}\times600=300\ K=27^\circ C$