Question

9. An ice tray contains 800 g of liquid water at 0°C. Calculate the change in entropy of the water as it
freezes slowly and completely at 0°C. The latent heat of fusion is L= 333 kJ/kg.
A. -975.8 J/K
B. -623.2 J/K
C.975.8 J/K
D. 623.2 J/K

Answer 1

The change in entropy of the water is : A. -975.8 J/K

Explanation:

Given that mass of liquid water is 800g at 0°C and the latent heat of fusion is L=333000J/Kg then the concept to apply here is;

Change in entropy, dS=dQ/T

This can be written as ΔS=ΔQ/T where you calculate change in entropy, the water changes phases while temperature remains constant.

Applying the formula;

ΔS=ΔQ/T =ΔQ= -mL where m is mass of water and L is latent heat of fusion =333000J/kg

ΔS= -(0.8 kg)*(333000 J/kg) /273 K = -975.8 J/K

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Keywords : liquid water, change in entropy, water, freezes slowly, latent heat of fusion

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