A Carnot freezer takes heat from water at $0 ° C$ inside it and rejects it to the room at a temperature of $27 ° C .$ The latent heat of ice is $336 \times 10^{3} J k g^{- 1} .$ If 5 kg of water at $0 ° C$ is converted into ice at $0 ° C$ by the freezer, then the energy consumed by the freezer is closed to

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$1.67 \times 10^{5} J$

$1.51 \times 10^{5} J$

$1.68 \times 10^{6} J$

$1.71 \times 10^{7} J$

answer is D.

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Detailed Solution

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By conservation of energy
$Q_{1} + W_{e x t} = Q_{2}$
$\Rightarrow W_{e x t} = Q_{2} - Q_{1}$
For a Carnot freezer we use
$\frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}} .... (1)$
Here $Q_{1} \to$ heat released by freezer in freezing ice
Here $Q_{1} = m L = 5 \times 336 \times 10^{3} (J / k g) \times k g$
We use by equation - (1)
$Q_{2} = (\frac{T_{2}}{T_{1}}) \times Q_{1} = \frac{300}{273} \times 5 \times 336 \times 10^{3}$
Energy consumed by freezer in
$W = Q_{2} - Q_{1}$
$\Rightarrow W = 5 \times 336 \times 10^{3} (\frac{300}{273} - 1)$
$\Rightarrow W = 1.67 \times 10^{5} J$