Question

A 43 g ice cube at -69°C is placed in a lake whose temperature is 81°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. (Hint: Will the ice cube affect the temperature of the lake?)

Answer 1

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

ΔS = Q/T
(note that Q is positive when heat is added and negative if heat is removed)

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 = mxCxln(T₂/T₁)

You can 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 = (-69 + 273)K = 204 K
to melting temperature
T₂ = 273 K
we know the heat of fusion of ice is
∆Hf = 334 kJkg⁻¹
After the melting process the molten ice heats up to the temperature of the lake
T₃ = (81 + 273) K = 354 K
The specific heat capacity of liquid water is:
C_w = 4182 Jkg⁻¹K⁻¹
C_i=2220j/kg.k
So the change entropy of the water in the ice cube is:
ΔS_cube = m_cubex[( C_ixln(T₂/T₁)] + (∆Hf/T₂) +( C_wxln(T₃/T₂)]
= 0.043 x[( 2220 xln(273/204) )+ ((334×10³)/273 ) + (4182 xln(354/273))]

=0.043x(646.8+1223.44+1086.588)

=127.14 jk^-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_cubex[( (C_ix(T₂ - T₁)) + ∆Hf + (C_wx(T₃ - T₂) )]
= 0.043 kg x[( 2220 x(273 - 204)) + (334×10³ )+ (4182 x(354 - 273) ]

=0.043x(153180+334000+338742)
= 35514.6 J
So the entropy of the lake changes by
Δ_lake = -Q/T₃
= - 35514.6 J / 354 K
= - 100.3 J∙K⁻¹

The total change in entropy of the cube lake system is
ΔS = ΔS_cube + Δ_lake
= 127.14 - 100.3
= 26.84 J∙K⁻¹