Question

1. 20 kg of 0 C liquid water is placed in a freezer. If the temperature remains at the freezing point, what is the change in entropy after it has frozen?1. 1.22 kJ/K2. -24.4 kJ/K3. 6.66x10^6 kJ/K4. Infinite

Answer 1

From the information given,

mass of water, m = 20 kg

Temperature of water, θ = 0 C

we would convert the temperature to kelvin by adding 273. Thus,

θ = 0 + 273 = 273 K

Latent heat of fusion of water, hf = 3.33 x 10^5 j/kg

In this scenario, phase change occurs at constant temperature. Thus,

total heat interaction during phase change is

dQ = mhf = 20 x 3.33 x 10^5 = 6.66 x 10^6 J

Change in entropy, dS = dQ/θ = 6.66 x 10^6/273

dS = 24.396 x 10^3 j/k

Converting to Kj/k, we would divide by 1000. Thus,

dS = 24.396 KJ/k

There would be a decrease in entropy since the phase transition is towards lower internal energy.Thus,

dS = - 24.4 KJ/k