Question

A 70.0 g ice cube at 0.0 degrees C is placed in a lake whose temperature is 18.0 degrees C. Calculate the change in entropy (in joules/Kelvin) of the system as the ice cube comes to thermal equilibrium with the lake. (c for water = 4186 J/kg-K)

Answer 1

Solution:

Assuming constant heat capacity, the change in entropy is given by

ΔS = m∙C∙ln(T₂/T₁)


Consider the ice cube. It heats up from
T₁ = (0 + 273)K = 273 K
to
T₂ = (18+273) = 291 K

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 + (334×10³ J∙kg⁻¹/273 K) + 4186 J∙kg⁻¹∙K⁻¹∙ln(291/273) ) * 0.07

= 104.33 J∙K⁻¹

Given, 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.

Q = m_cube∙( Ci∙(T₂ - T₁) + ∆Hf + Cw∙(T₃ - T₂) )
= 0.07 kg ∙( 2220 J∙kg⁻¹∙K⁻¹∙(273K - 273K) + 334×10³ J∙kg⁻¹ + 4186 J∙kg⁻¹∙K⁻¹∙(291K - 273K) )
= 98656 J

So the entropy of the lake changes by:
ΔS_lake = -Q/T₃
= - 98656 J / 291 K
= - 339.02 J∙K⁻¹

The total change in entropy of the cube lake system is:
ΔS = ΔS_cube + ΔS_lake
= 104.33 J∙K⁻¹ - 339.02 J∙K⁻¹
= 234.68 J∙K⁻¹