A Carnot engine works between ice point and steam point. In order to increase the efficiency of the engine by 20%, its source temperature will have to changed by

29.5K Increase

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29.5 K decrease

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{37.5}^{0} C

increase

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{29.5}^{0} C

decrease

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Solution

The correct option is A 29.5K Increase
Efficiency of a Carnot engine $(η) = 1 - \frac{T_{c}}{T_{h}}$
$η = 1 - \frac{273}{373} = 1 - 0.7319 = 0.2681$
Now to increase the efficiency by 20% ; i.e. $0.2 \times 0.2681 = 0.0536$ ;
New efficiency $= 0.2681 + 0.0536 = 0.3217$
$0.3217 = 1 - \frac{273}{T_{h}}$
$\frac{273}{T_{h}} = 0.6783$
$T_{h} = \frac{273}{0.6783} = 402.48 K$
Change in source temperature $= 402.48 - 373 = 29.48 K$

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