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A carnot engine whose low temperature reservoir is at $7^{0}$ 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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280K

373.3K

560K

840 K

answer is D.

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Detailed Solution

$η = 1 - \frac{T_{2}}{T_{1}}$

$0.5 = \frac{T_{1} - (273 + 7)}{T_{1}}$

$T_{1} = 560 K$

$η^{1} = \frac{T_{1}^{1} - (273 + 7)}{T_{1}^{1}}$

$0.7 = \frac{T_{1}^{1} - (280)}{T_{1}^{1}}$

$T_{1}^{1} = 933 K$

$Δ T = 933 - 560$

=373.3K

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