The output power of the motor of a refrigerator is 240 W . The te… - Vidyadip

Question

The output power of the motor of a refrigerator is 240 W . The temperature of the freezing chamber is 240 K and the outside air is 300 K . How much heat can be removed from the freezer in 10 minutes? Assuming that there is ideal efficiency, what is the minimum time in which 10 kg of water at 273 K can be converted in to ice? $J =4.2 \times 10^3 JK cal ^{-1}$.

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Answer

We know that :
$\alpha=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}$
Here given :
$\begin{aligned}T_1 & =300 K \\T_2 & =240 K \\W & =240 W=240 J / s \\Q_1 & =? \\t & =?\end{aligned}$
$\begin{aligned}\therefore \quad Q_2 & =W(\alpha)=\frac{W \cdot T_2}{T_1-T_2} \\& =240 \times \frac{240}{300-240} \\& =240 \times \frac{240}{60}=240 \times 4=960 J / 5\end{aligned}$
(i) Suppose heat removed from freezing chamber in 10 minutes $= Q$.
$\begin{aligned}\therefore \quad Q & =Q_2 \times 10 min . \\& =960 \times 10 \times 60=576000 J \\Q & =576000 J \\& =\frac{576000}{4.2 \times 10^3}=137.14 Kcal \\& {\left[\because J=4.2 \times 10^3 J / K Cal\right] }\end{aligned}$
(ii) To convert 1 kg of water into ice at 273 K desired heat
$\begin{aligned}Q & =m \times L=1 \times 80 Kcal \\& =80 \times 4.2 \times 10^3 J\end{aligned}$
Suppose the heat extracted in time t is $Q ^{\prime}$.
$\therefore$ Rate of heat removal from freezing chamber
$=\frac{80 \times 4.2 \times 10^3}{t} J / s$
This rate should be equal to $Q_2$
i.e.
$\begin{aligned}960 & =\frac{80 \times 4.2 \times 10^3}{t} \\t & =\frac{80 \times 4.2 \times 10^3}{960} \\& =\frac{8 \times 42}{96} \times 10^2=\frac{336}{96} \times 10^2 \\t & =3.5 \times 10^2{ }_5=350s\end{aligned}$

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