The ratio ($W/Q$) for a carnot-engine is $1/6$. Now the temperature of sink is reduced by $62\text{ }^\circ\text{C}$, this ratio becomes twice, therefore the initial temp. of the sink and source are respectively: (2000)
$33\text{ }^\circ\text{C}$, $67\text{ }^\circ\text{C}$
$37\text{ }^\circ\text{C}$, $99\text{ }^\circ\text{C}$
$67\text{ }^\circ\text{C}$, $33\text{ }^\circ\text{C}$
$97\text{ K}$, $37\text{ K}$
Solution:
Ratio $W/Q_1$ is the efficiency $\eta_1 = 1/6$. When sink temperature decreases, efficiency becomes $2/6 = 1/3$. Solving yields source $T_1 = 372\text{ K} = 99\text{ }^\circ\text{C}$ and sink $T_2 = 310\text{ K} = 37\text{ }^\circ\text{C}$.
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