Question

At very low temperatures the molar specific heat C_V for many solids is (approximately) proportional to T³; that is, C_V=AT³, where A depends on the particular substance. For aluminum, A=3.155×10^-5 J/mol·K⁴. Find the entropy change of 3.1 moles moles of aluminum when its temperature is raised from 5.0 to 9.8 K.

Answer 1

The formula for an entropy change is the following:

dS = dQ/T

This means infinitesimal addition of heat divided by the temperature at which the heat destination currently reads gives the infinitesimal addition of entropy.

How much infinitesimal heat is added at any given temperature?
dQ = N*cv * (T + dT - T)

Cancel the T's:
dQ = N*cv(T) * dT

Update relation for entropy change:
dS = N*cv(T) * dT/T

Integrate both sides:
∆S = N*∫ cv(T)/T dT

Plug in expression for cv(T) = A*T^3
∆S = N*∫ (A*T^3)/T dT

Simplify:
∆S = N*A*∫ T^2 dT

Temperatures are in Kelvin, so we are OK.

∆S = ⅓N*A*(T2^3 - T1^3)

Data:
N:= 3.1 moles; A:=3.155e-5 J/mole-K^4; T1:=5 K; T2:=9.8 K;

∆S = ⅓N*A*(T2^3 - T1^3)
= 1/2*3.1*3.155*10^-5(9.8^2-5^2)
∆S = 0.003474 Joules/K