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Conduction Law

A refrigirato...

Question

A refrigirator converts $100 g$ of water at $20^{o} C$ to ice at $- 10^{o} C$ in $35$ min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is $4.2 J g^{- 1}$ $K^{- 1}$ , specific heat of ice is $336 J g^{- 1}$ and the specific heat capacity of ice is $2.1 J g^{- 1}$ $K^{- 1}$ .

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Solution

Mass of water $= 100 g = 0.1 k g$
The temperature of water $= 20^{\circ} C$

Amount of heat extracted to convert water from $20^{\circ} C$ to $0^{\circ} C$ is,
$Q_{1} = m C_{w a t e r} Δ t = 0.1 \times 4200 \times (20 - 0) = 8400 J$
Amount of heat extracted to convert water at $0^{\circ} C$ to $0^{\circ} C$ ice is,
$Q_{2} = m L_{i c e} = 0.1 \times 336 \times 1000 = 33600 J$
Amount of heat extracted to convert ice at $0^{\circ} C$ to ice at $- 10^{\circ} C$ is,
$Q_{3} = m L_{i c e} Δ t = 0.1 \times 2.1 \times 10^{3} \times (0 - (- 10)) = 21000 J$

Total heat ( $Q$ ) $= Q_{1} + Q_{2} + Q_{3}$
$= 8400 + 33600 + 2100$
$= 44100 J$

Therefore,
$Q = P t$
$P = \frac{Q}{t}$
Power $= \frac{44100}{35 \times 60} = 21 W$

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