An engine has an efficiency of $0.25$ when temperature of sink is reduced by $58\,^oC$, if its efficiency is doubled, then the temperature of the source is ..... $^oC$
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Here, $\eta_{1}=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}$

$0.25=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}} \Rightarrow \frac{1}{4}=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}$

$\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=1-\frac{1}{4}=\frac{3}{4}$   $ . .(\mathrm{i})$

According to question.

$\eta_{2}=2 \eta_{1},$ and $\mathrm{T}_{2}=\mathrm{T}_{2}-58^{\circ} \mathrm{C}$

$\therefore \quad 2 \times \frac{1}{4}=1-\frac{\left(\mathrm{T}_{2}-58^{\circ} \mathrm{C}\right)}{\mathrm{T}_{1}}$

$\Rightarrow 1-\frac{1}{2}=\frac{\mathrm{T}_{2}-58^{\circ} \mathrm{C}}{\mathrm{T}_{1}}$

$\frac{1}{2}=\frac{T_{2}}{T_{1}}-\frac{58^{\circ}}{T_{1}} \Rightarrow \frac{3}{4}-\frac{1}{2}=\frac{58}{T_{1}}$

$\Rightarrow \mathrm{T}_{1}=232^{\circ} \mathrm{C}$

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