JEE Main 2016 · 10 Apr · Shift 1 · Physics Q8

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PhysicsSingle correct+4 / -1Question 8 of 90

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 $336\times10^3$ J kg $^{-1}$ . 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 close to :

1.67\times10^5

1.68\times10^6

1.51\times10^5

1.71\times10^7

Answer and worked solution

Official answer: A

Heat removed: $Q_c=5\times336\times10^3=1.68\times10^6$ J. For a Carnot freezer,
$W=Q_c\frac{T_h-T_c}{T_c}=1.68\times10^6\times\frac{27}{273}\approx1.67\times10^5\text{ J}.$

Topics:Carnot cycle Latent heat and phase change

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JEE Main 2016 · 10 Apr · Shift 1, Physics question 8