Temperature of a cold reservoir of a Carnot engine is $127^{\circ} \mathrm{C}$. If the efficiency of the…

Temperature of a cold reservoir of a Carnot engine is $127^{\circ} \mathrm{C}$. If the efficiency of the Carnot engine is $20 \%$, then the temperature of the hot reservoir is
  1. $500^{\circ} \mathrm{C}$
  2. $227^{\circ} \mathrm{C}$
  3. $273^{\circ} \mathrm{C}$
  4. $400^{\circ} \mathrm{C}$

Solution

As $\eta=1-\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{T}_{\mathrm{H}}} \Rightarrow 0.2=1-\frac{400}{\mathrm{~T}_{\mathrm{H}}}$ $\begin{aligned} & \Rightarrow \frac{400}{\mathrm{~T}_{\mathrm{H}}}=0.8 \\ & \Rightarrow \mathrm{T}_{\mathrm{H}}=500 \mathrm{~K}=227^{\circ} \mathrm{C}\end{aligned}$

Asked in: AP EAMCET 2022 (08 Jul Shift 1)

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