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A Carnot engine whose low-temperature reservoir is at 7° C has an efficiency of 50%. It is desired to increase the efficiency to 70%. By how many degrees should the temperature of the high temperature reservoir be increased

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840 K

280 K

560 K

380 K

answer is D.

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Detailed Solution

Initially $η= \frac{T_{1} - T_{2}}{T_{1}}$

$\Rightarrow 0 .5 = \frac{T_{1} - (273 + 7)}{T_{1}}$

$\Rightarrow \frac{1}{2} = \frac{T_{1} - 280}{T_{1}} \Rightarrow T_{1} = 560 K$

Finally $η_{1}^{'} = \frac{T_{1}^{'} - T_{2}}{T_{1}^{'}} \Rightarrow 0 .7 = \frac{T_{1}^{'} - (273 + 7)}{T_{1}^{'}}$

$\Rightarrow T_{1}^{'} = 933 K$

$∴$ increase in temperature

$= 933 - 560 = 373 K \approx 380 K$

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