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The efficiency of Carnot engine is 50% and temperature of sink is 500K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of the sink will be : -
A
100 K
B
600 K
C
400 K
D
500 K
Detailed Solution
Efficiency of a Carnot engine: $\eta = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the source and $T_2$ the sink temperature.
For 50 percent: $0.5 = 1 - \dfrac{500}{T_1} \Rightarrow T_1 = 1000$ K
For 60 percent with the same source: $0.6 = 1 - \dfrac{T_2}{1000}$
$\Rightarrow T_2 = 0.4 \times 1000 = 400$ K
So the sink temperature must be lowered to 400 K.
For 50 percent: $0.5 = 1 - \dfrac{500}{T_1} \Rightarrow T_1 = 1000$ K
For 60 percent with the same source: $0.6 = 1 - \dfrac{T_2}{1000}$
$\Rightarrow T_2 = 0.4 \times 1000 = 400$ K
So the sink temperature must be lowered to 400 K.