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If $\sin k$ is at a temperature of $-39^{\circ} \mathrm{C}$ and source at $0^{\circ} \mathrm{C}$, then efficiency will be
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1854 Upvotes
Verified Answer
The correct answer is:
$14.2 \%$
Given, temperature of $\operatorname{sink}$,
$$
\begin{aligned}
T_{2} &=-39^{\circ} \mathrm{C} \\
&=(273-39) \mathrm{K} \\
&=234 \mathrm{~K}
\end{aligned}
$$
Temperature of source,
$$
\begin{aligned}
T_{1} &=0^{\circ} \mathrm{C} \\
&=(273+0) \mathrm{K} \\
&=273 \mathrm{~K} \\
\therefore \text { Efficiency, } \eta &=1-\frac{T_{2}}{T_{1}} \\
&=1-\frac{234}{273} \\
&=\frac{39}{273} \\
&=0.142 \\
&=14.2 \%
\end{aligned}
$$
$$
\begin{aligned}
T_{2} &=-39^{\circ} \mathrm{C} \\
&=(273-39) \mathrm{K} \\
&=234 \mathrm{~K}
\end{aligned}
$$
Temperature of source,
$$
\begin{aligned}
T_{1} &=0^{\circ} \mathrm{C} \\
&=(273+0) \mathrm{K} \\
&=273 \mathrm{~K} \\
\therefore \text { Efficiency, } \eta &=1-\frac{T_{2}}{T_{1}} \\
&=1-\frac{234}{273} \\
&=\frac{39}{273} \\
&=0.142 \\
&=14.2 \%
\end{aligned}
$$
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