At fixed volume, a reversible infinitesimal heating supplies

δ Q_{rev} = C_{V} d T

. Since entropy satisfies

d S = δ Q_{rev} / T

Δ S = C_{V} \int_{T_{i}}^{T_{f}} \frac{d T}{T} = C_{V} lo g \frac{T _{f}}{T _{i}} .

ii

Solution

Let the common final temperature be

T_{f}

. A reversible isolated process has zero total entropy production, so part (i) gives

C_{V} lo g \frac{T _{f}}{T _{1}} + C_{V} lo g \frac{T _{f}}{T _{2}} = 0.

Hence

T_{f} = T_{1} T_{2}

for both bodies. The decrease in their total internal energy is extracted as work by the reversible engine mediating the transfer.

Direct contact permits no work extraction, so conservation of internal energy gives

C_{V} (T_{f} - T_{1}) + C_{V} (T_{f} - T_{2}) = 0, T_{f} = \frac{T _{1} + T _{2}}{2} .

The total entropy change is

Δ S = C_{V} lo g \frac{T _{f}}{T _{1}} + C_{V} lo g \frac{T _{f}}{T _{2}} = C_{V} lo g \frac{( T _{1} + T _{2} ) ^{2}}{4 T _{1} T _{2}} \geq 0.

The arithmetic-geometric mean inequality makes the last expression positive when

T_{1} \neq = T_{2}

, with equality only when the bodies were already at the same temperature. This is the entropy production of irreversible heat flow.

There are no discussions about this article yet.

There are currently no matching articles.