How much energy in watt hour may be required to convert 2 kg of water into ice at 0°C, assuming that the refrigerator is ideal? Given temperature of freezer is -15°C, room temperature is25°C and initial temperature of water is 25°C.

Question

How much energy in watt hour may be required to convert 2 kg of water into ice at 0°C, assuming that the refrigerator is ideal? Given temperature of freezer is -15°C, room temperature is25°C and initial temperature of water is 25°C. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Here T1 = 273 + 25 = 298 K and T2 = 273 - 15 = 258 K Specific heat of water, C = 4.2 × 103 J/kg K Latent heat of fusion of ice, L= 3.36 × 105 J/kg The amount of heat required to transform water of 25°C into ice of 0°C Q2 = mCΔ T +mL = 2 × 4.2 × 103 × (25 -0) + 2× 3.36 × 105 = 2.1 × 105 + 6.72 × 105 = 8.82 × 105J Heat rejected to the surroundings; we have (Q1/Q2) = (T1/T2) ∴ Q1 = Q2 ((T1/T2)) = 8.82 ×105 ((298/258)) = 10.19 ×105 J Energy supplied to convert water into ice, W = Q1 - Q2 = 10.15 × 105 -8 .8 2 × 105 = 1.33 × 105 J = (1.33 ×105/3600) = 36.96 Wh

Q. How much energy in watt hour may be required to convert 2 kg of water into ice at 0∘C, assuming that the refrigerator is ideal? Given temperature of freezer is −15∘C, room temperature is25∘C and initial temperature of water is 25∘C.

Q. How much energy in watt hour may be required to convert $2$ kg of water into ice at $0^{\circ} C$ , assuming that the refrigerator is ideal? Given temperature of freezer is $- 1 5^{\circ} C$ , room temperature is $2 5^{\circ} C$ and initial temperature of water is $2 5^{\circ} C$ .