Question
Heat is transferred from inside a refrigerator at 287 K to a room at 310 K ideally, how much heat (in joules) will be given to the room by the expenditure of each joule of electrical energy?
✓
Answer
Given that :
$\begin{aligned}T_2 & =287 K \\T_1 & =310 K \\Q_1 & =? \\W & =1 J\end{aligned}$
$\begin{array}{l}\text {Using the relation :}\quad\alpha=\frac{T_2}{T_1-T_2} \\\alpha=\frac{287}{310-287}=\frac{287}{23} \\\text {We get}\quad\quad\alpha=12.48\end{array}$
$\begin{array}{l}\text {We also know that}\quad\alpha=\frac{Q_2}{W}=\frac{Q_2}{1} \\\alpha=Q_2\end{array}$
$\therefore \quad Q_2=12.48 J$
$\text {$\therefore$ Now}\quad W=Q_1-Q_2$
$\text {or}\quad Q_1=W+Q_2$
Total heat supplied to the room :
$\begin{array}{l}Q_1=1+12.48 \\Q_1=13.48 J\end{array}$
$\begin{aligned}T_2 & =287 K \\T_1 & =310 K \\Q_1 & =? \\W & =1 J\end{aligned}$
$\begin{array}{l}\text {Using the relation :}\quad\alpha=\frac{T_2}{T_1-T_2} \\\alpha=\frac{287}{310-287}=\frac{287}{23} \\\text {We get}\quad\quad\alpha=12.48\end{array}$
$\begin{array}{l}\text {We also know that}\quad\alpha=\frac{Q_2}{W}=\frac{Q_2}{1} \\\alpha=Q_2\end{array}$
$\therefore \quad Q_2=12.48 J$
$\text {$\therefore$ Now}\quad W=Q_1-Q_2$
$\text {or}\quad Q_1=W+Q_2$
Total heat supplied to the room :
$\begin{array}{l}Q_1=1+12.48 \\Q_1=13.48 J\end{array}$
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