Question

A 73 g ice cube at -45°C is placed in a lake whose temperature is 76°C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg·K.

Answer 1

Change in entropy associated with transfer of amount Q of heat at constant thermodynamic temperature T is

ΔS = Q/T

Assuming constant heat capacity the change in entropy associated with a process in which the temperature changes from T₁ to T₂, is given by
ΔS = m∙C∙ln(T₂/T₁)

We assume that mass of the ice is small compared to the amount of water in the lake. Therefore the change in temperature of the lake water due the melting of the cube is negligibly small.

Consider the ice cube. It heats up from
T₁ = (-45 + 273)K = 228 K
to melting temperature
T₂ = 273 K
Then it absorbs the heat of fusion
∆Hf = 334.0 kJ∙kg⁻¹
After the melting process the molten ice heats up to the temperature of the lake
T₃ = (76 + 273) K = 349 K
The specific heat capacity of liquid water is:
Cw = 4182 J∙kg⁻¹∙K⁻¹

So the change entropy of the water in the ice cube is:
ΔS_cube = m_cube∙( Ci∙ln(T₂/T₁) + (∆Hf/T₂) + Cw∙ln(T₃/T₂) )
= 0.073 kg ∙( 2220 J∙kg⁻¹∙K⁻¹∙ln(273/228) + (334×10³ J∙kg⁻¹//273 K) + 4182 J∙kg⁻¹∙K⁻¹∙ln(349/273) )
= 0.073*( 399.88 + 1223.44 + 1027.10) = 193.48 J∙K⁻¹

As pointed above the lake's temperature is constant. The amount of heat removed from the lake equals the amount of heat absorbed from the cube, as it heats up and melts.
This amount is:
Q = m_cube∙( Ci∙(T₂ - T₁) + ∆Hf + Cw∙(T₃ - T₂) )
= 0.073 kg ∙( 2220 J∙kg⁻¹∙K⁻¹∙(273K - 228K) + 334×10³ J∙kg⁻¹ + 4182 J∙kg⁻¹∙K⁻¹∙(349K - 273K) )
= 0.073*( 99900 + 334000 + 317832) = 54876.44 J

So the entropy of the lake changes by:

ΔS_lake = -Q/T₃
= - 54876.44 J / 349 K
= - 157.24 J∙K⁻¹

The total change in entropy of the cube lake system is:
ΔS = ΔS_cube + ΔS_lake
= 193.48 J∙K⁻¹ - 157.24 J∙K⁻¹
= 36.24 J∙K⁻¹