A Carnot engine has efficiency of $50 \%$. If the temperature of sink is reduced by $40^{\circ} \mathrm{C}$, its efficiency increases by $30 \%$. The temperature of the source will be:
Solution
<p>$1 - {{{T_L}} \over {{T_H}}} = 0.5$ ...... (1)</p>
<p>$1 - {{{T_L} - 40} \over {{T_H}}} = 0.65$ .... (2)</p>
<p>$\Rightarrow {T_H} = {{800} \over 3}\,K \simeq 266.7\,K$</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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