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A Carnot engine takes $3 \times 10^6$ calories of heat from reservoir at $627^{\circ} \mathrm{C}$ and gives…
A Carnot engine takes $3 \times 10^6$ calories of heat from reservoir at $627^{\circ} \mathrm{C}$ and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is
zero $8.4 \times 10^6 \mathrm{~J}$ $4.2 \times 10^6 \mathrm{~J}$ $16.8 \times 10^6 \mathrm{~J}$ Solution
Heat taken by Carnot engine,
$
Q=3 \times 10^6 \mathrm{cal}=4.2 \times 3 \times 10^6 \mathrm{~J}
$
Temperature of reservoir (source)
$
T_1=(627+273) \mathrm{K}=900 \mathrm{~K}
$
Temperature of sink,
$
T_2=(27+273) \mathrm{K}=300 \mathrm{~K}
$
We know that, efficiency of Carnot engine,
$
\boldsymbol{\eta}=\frac{W}{Q}=\mathbf{l}-\frac{T_2}{T_1}
$
$
\begin{aligned}
& \Rightarrow \quad \frac{W}{Q}=1-\frac{T_2}{T_1} \\
& \Rightarrow \quad \frac{W}{42 \times 3 \times 10^6}=1-\frac{300}{900} \\
& \Rightarrow \quad \frac{W}{42 \times 3 \times 10^6}=\frac{2}{3} \\
& \Rightarrow \quad W=\frac{2}{3} \times 4.2 \times 3 \times 10^6=8.4 \times 10^6 \mathrm{~J}
\end{aligned}
$
Asked in: AP EAMCET 2020 (22 Sep Shift 1)
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