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A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by 62 $\mathrm{K}$, its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ are, respectively :
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The correct answer is:
$372 \mathrm{~K}$ and $310 \mathrm{~K}$
$372 \mathrm{~K}$ and $310 \mathrm{~K}$
$\eta_1=\frac{T_1-T_2}{T_1}=\frac{1}{6}$
$\eta_2=\frac{T_1-\left(T_2-62\right)}{T_1}=\frac{1}{3}$
$\Rightarrow \frac{T_1-T_2}{T_1}+\frac{62}{T_1}=\frac{1}{3}$
$\frac{1}{6}+\frac{62}{T_1}=\frac{1}{3}$
$\frac{62}{T_1}=\frac{1}{6}$
$\therefore T_1=62 \times 6=372 \mathrm{~K}$
$\frac{T_1-T_2}{T_1}=\frac{1}{6}$
$1-\frac{T_2}{T_1}=\frac{1}{6}$
$\frac{T_2}{372}=\frac{5}{6}$
$\Rightarrow T_2=310 \mathrm{~K}$
$\eta_2=\frac{T_1-\left(T_2-62\right)}{T_1}=\frac{1}{3}$
$\Rightarrow \frac{T_1-T_2}{T_1}+\frac{62}{T_1}=\frac{1}{3}$
$\frac{1}{6}+\frac{62}{T_1}=\frac{1}{3}$
$\frac{62}{T_1}=\frac{1}{6}$
$\therefore T_1=62 \times 6=372 \mathrm{~K}$
$\frac{T_1-T_2}{T_1}=\frac{1}{6}$
$1-\frac{T_2}{T_1}=\frac{1}{6}$
$\frac{T_2}{372}=\frac{5}{6}$
$\Rightarrow T_2=310 \mathrm{~K}$
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