Question
A Carnot engine with sink's temperature at $17\,^oC$ has $50\%$ efficiency. By how much should its source temperature be changed to increases its efficiency to $60\%$ ?...... $K$
✓
Answer
$\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}$
$\Rightarrow \frac{50}{100}=1-\frac{290}{\mathrm{T}_{1}} \Rightarrow \mathrm{T}_{1}=580 \mathrm{K}$
$\Rightarrow \frac{60}{100}=1-\frac{290}{\mathrm{T}_{1}^{\prime}} \Rightarrow \mathrm{T}_{1}^{\prime}=725 \mathrm{K}$
$\Rightarrow \Delta \mathrm{T}=\mathrm{T}_{1}^{\prime}-\mathrm{T}_{1}=145 \mathrm{K}$
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