First slide

Refrigerator

Question

A cannot engine extracts heat from water at 0°C and rejects it to room at 24.4°C. The work required by the refrigerator for every 1 kg of water converted into ice (latent heat of ice = 336kj/kg ) is

Easy

A

30 KJ
B

336 KJ
C

11.2 KJ
D

24.4 KJ

Solution

T₂ = 273k, T₁ = 297.4k
w = ? m–1kg
L_ice = 336000J/kg
$COP = \frac{{{Q_2}}}{w} = \frac{{{T_2}}}{{{T_1} - {T_2}}} \Rightarrow \frac{{m{L_{ice}}}}{W} = \frac{{{T_2}}}{{{T_1} - {T_2}}}$
$\frac{{1 \times 336000}}{w} = \frac{{273}}{{24.4}}$
$w = \frac{{336000 \times 24.4}}{{273}} = 30.03 \times {10^3}J = 30kJ$

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