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A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency :-
A
275 K
B
325 K
C
250 K
D
380 K
Detailed Solution
Efficiency of a Carnot engine: $\eta = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ the sink temperature.
$1 - \dfrac{300}{T_1} = 0.4 \Rightarrow \dfrac{300}{T_1} = 0.6 \Rightarrow T_1 = 500$ K
The efficiency is to be increased by 50 percent of the original efficiency: $\eta' = 0.4 + 0.4 \times \dfrac{50}{100} = 0.6$
Let the source temperature be increased by $\Delta T$, the sink remaining at 300 K.
$0.6 = 1 - \dfrac{300}{500 + \Delta T}$
$\dfrac{300}{500 + \Delta T} = 0.4 \Rightarrow 500 + \Delta T = 750$
$\Rightarrow \Delta T = 250$ K
$1 - \dfrac{300}{T_1} = 0.4 \Rightarrow \dfrac{300}{T_1} = 0.6 \Rightarrow T_1 = 500$ K
The efficiency is to be increased by 50 percent of the original efficiency: $\eta' = 0.4 + 0.4 \times \dfrac{50}{100} = 0.6$
Let the source temperature be increased by $\Delta T$, the sink remaining at 300 K.
$0.6 = 1 - \dfrac{300}{500 + \Delta T}$
$\dfrac{300}{500 + \Delta T} = 0.4 \Rightarrow 500 + \Delta T = 750$
$\Rightarrow \Delta T = 250$ K