Worked examples · Sheet 3

Changes in entropy in heat transfers

Theory used: Entropy of reservoirs, bodies and changes of state Entropy Principle of entropy increase

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A heat of $2.0 kJ$ passes by conduction, through a metal rod, from a reservoir at $400 K$ to a reservoir at $250 K$ ; the state of the rod does not change. Compute the change in entropy of each reservoir and of the universe.

T_{1} = 400 K, T_{2} = 250 K, Q = 2.0 kJ

Δ S_{1} = \frac{- Q}{T _{1}}, Δ S_{2} = \frac{Q}{T _{2}}, Δ S_{u} = Q (\frac{1}{T _{2}} - \frac{1}{T _{1}})

Δ S_{1} = - \frac{2000 J}{400 K} = - 5.0 J/K, Δ S_{2} = \frac{2000 J}{250 K} = 8.0 J/K

Δ S_{u} = - 5.0 J/K + 8.0 J/K = 3.0 J/K

A reservoir exchanges heat at constant temperature, therefore its change in entropy is the heat absorbed divided by its temperature. The hot reservoir absorbs $- 2.0 kJ$ and its entropy decreases; the cold reservoir absorbs $+ 2.0 kJ$ and its entropy increases by more, because the same heat is divided by a lower temperature. The rod returns to the same state, therefore its entropy does not change, and the universe consists of the two reservoirs. The entropy of the universe increases by $3.0 J/K$ , in agreement with the principle of entropy increase, since conduction from a hot body to a cold one is an irreversible process.

Exercise 2

$2.0 kg$ of water is heated from $20^{\circ} C$ to $80^{\circ} C$ . Compute the change in entropy of the water, taking the specific heat $c = 4186 J/ (kg \cdot K)$ as constant.

m = 2.0 kg, T_{A} = 293 K, T_{B} = 353 K

d S = \frac{δ Q}{T} = \frac{m c d T}{T} ⟹ Δ S = \int_{T_{A}}^{T_{B}} \frac{m c d T}{T} = m c ln \frac{T _{B}}{T _{A}}

Δ S = 2.0 kg \cdot 4186 J/ (kg \cdot K) \cdot ln \frac{353}{293} \approx 8372 J/K \cdot 0.186 \approx 1.56 \cdot 1 0^{3} J/K

Entropy is a state function, therefore its change is computed along a reversible process between the initial state and the final one, whatever the way in which the water is heated. Consider heating the water with a succession of reservoirs at increasing temperatures, each one infinitesimally hotter than the water; in each step the water absorbs $δ Q = m c d T$ at the temperature $T$ , and the integral of $\frac{d T}{T}$ is the logarithm of the ratio of the temperatures, which must be expressed in kelvins. The result is positive, since the water absorbs heat. As a check, the heat absorbed is $Q = m c Δ T \approx 5.02 \cdot 1 0^{5} J$ , and divided by the average temperature $323 K$ it gives about $1.55 \cdot 1 0^{3} J/K$ , a value close to the exact one because the temperature interval is small compared with the absolute temperature.

Exercise 3

A block of ice of mass $0.50 kg$ at $0^{\circ} C$ melts completely in a room at $25^{\circ} C$ , which behaves as a reservoir; the resulting water stays at $0^{\circ} C$ . With the latent heat of fusion $λ_{f} = 3.33 \cdot 1 0^{5} J/kg$ , compute the change in entropy of the ice, of the room and of the universe.

m = 0.50 kg, T_{g} = 273 K, T_{s} = 298 K

Q = m λ_{f}, Δ S_{g} = \frac{m λ _{f}}{T _{g}}, Δ S_{s} = - \frac{m λ _{f}}{T _{s}}

Q = 0.50 kg \cdot 3.33 \cdot 1 0^{5} J/kg = 1.665 \cdot 1 0^{5} J

Δ S_{g} = \frac{1.665 \cdot 1 0 ^{5} J}{273 K} \approx 610 J/K, Δ S_{s} = - \frac{1.665 \cdot 1 0 ^{5} J}{298 K} \approx - 559 J/K

Δ S_{u} \approx 610 J/K - 559 J/K = 51 J/K

During melting the temperature of the ice stays at $273 K$ , therefore its change in entropy is the latent heat absorbed divided by this temperature. The room releases the same heat at the constant temperature of $298 K$ , and its entropy decreases by a smaller amount, because the denominator is larger. The sum is positive, therefore the entropy of the universe increases by about $51 J/K$ . Melting in a warmer room is indeed irreversible, since heat passes between bodies at different temperatures.

Exercise 4

A copper block of mass $1.0 kg$ and specific heat $c = 385 J/ (kg \cdot K)$ , at a temperature of $400 K$ , is immersed in a lake at $300 K$ and reaches thermal equilibrium. Compute the change in entropy of the copper, of the lake and of the universe.

m = 1.0 kg, T_{A} = 400 K, T_{L} = 300 K

Δ S_{Cu} = m c ln \frac{T _{L}}{T _{A}}, Q_{L} = m c (T_{A} - T_{L}), Δ S_{L} = \frac{Q _{L}}{T _{L}}

Δ S_{Cu} = 385 J/K \cdot ln \frac{300}{400} \approx - 111 J/K

Q_{L} = 385 J/K \cdot 100 K = 3.85 \cdot 1 0^{4} J, Δ S_{L} = \frac{3.85 \cdot 1 0 ^{4} J}{300 K} \approx 128 J/K

Δ S_{u} \approx - 110.8 J/K + 128.3 J/K \approx 17.6 J/K

The lake has a very large heat capacity and behaves as a reservoir, therefore the copper cools down to $300 K$ and the temperature of the lake does not change. The change in entropy of the copper is computed as for a body with constant specific heat, along a reversible cooling between the same states, and it is negative because the copper releases heat. The lake absorbs the heat released by the copper, $m c (T_{A} - T_{L})$ , at the constant temperature of $300 K$ . The sum of the two changes is positive, therefore the entropy of the universe increases by about $17.6 J/K$ . The sign is the expected one, since cooling in contact with a colder body is irreversible.