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If heat engine is filled at temperature $27^{\circ} \mathrm{C}$ and heat of $100 \mathrm{k} \mathrm{cal}$ is taken from source at temperature $677^{\circ} \mathrm{C}$. Work done (in $\mathrm{J}$ ) is
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Verified Answer
The correct answer is:
$0.28 \times 10^6$
Given, $T_2=27^{\circ} \mathrm{C}=300 \mathrm{~K}$
$$
\begin{array}{rlrl}
& T_1 & =677^{\circ} \mathrm{C}=950 \mathrm{~K} \\
& \quad \eta & =\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1} \\
& =1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1} \\
\therefore \quad & \quad \eta & =1-\frac{300}{950}=\frac{13}{19} \\
\therefore \quad & W & =H Q_1=100 \times 90^3 \times \frac{13}{19} \mathrm{cal} \\
\text { Also, } 4.2 \mathrm{~J} & =1 \mathrm{cal} \\
\therefore \quad & W & =100 \times 10^3 \times \frac{13}{19} \times 4.2 \\
\Rightarrow \quad & W & =2.87 \times 10^5=0.28 \times 10^6 \mathrm{~J}
\end{array}
$$
$$
\begin{array}{rlrl}
& T_1 & =677^{\circ} \mathrm{C}=950 \mathrm{~K} \\
& \quad \eta & =\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1} \\
& =1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1} \\
\therefore \quad & \quad \eta & =1-\frac{300}{950}=\frac{13}{19} \\
\therefore \quad & W & =H Q_1=100 \times 90^3 \times \frac{13}{19} \mathrm{cal} \\
\text { Also, } 4.2 \mathrm{~J} & =1 \mathrm{cal} \\
\therefore \quad & W & =100 \times 10^3 \times \frac{13}{19} \times 4.2 \\
\Rightarrow \quad & W & =2.87 \times 10^5=0.28 \times 10^6 \mathrm{~J}
\end{array}
$$
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