Entropy and the direction of change

Which way?

The first law says energy is kept. It would allow heat to flow from a cold cup into a hot room, or a broken glass to jump back together. These never happen. We need a quantity that says which way change goes by itself. It is the entropy, $S$.

How entropy changes

When a body takes in heat $Q$ slowly at a steady temperature $T$ (in kelvin), its entropy rises by

The unit is J/K. Heat given out lowers it. Two useful results follow.

• A gas expanding at a steady temperature: $\Delta S = nR\ln\dfrac{V_2}{V_1}$.
• A body warmed from $T_1$ to $T_2$: $\Delta S = mc\ln\dfrac{T_2}{T_1}$.

Entropy depends only on the state, like internal energy. So for a sudden change we find $\Delta S$ along any slow path between the same two states.

The second law again

The total entropy of an isolated system never falls. It rises in every irreversible process and stays the same in a reversible one.

When heat $Q$ flows from hot to cold, the hot body loses $\dfrac{Q}{T_h}$ and the cold one gains $\dfrac{Q}{T_c}$, which is more. The total rises, so the flow goes that way and never back.

You should find that the total always rises, and that it rises less when the two waters are close in temperature. With no gap at all, nothing changes.

Watch out. One body can lose entropy, as the hot water does. Only the total of an isolated system cannot fall.

A worked example

$0.60\ \text{kg}$ of water at $90\ ^\circ\text{C}$ is mixed with $0.40\ \text{kg}$ at $15\ ^\circ\text{C}$. Find the change in the total entropy. ($c = 4186\ \text{J kg}^{-1}\text{K}^{-1}$)

1. Final temperature. The masses add to $1.00\ \text{kg}$, so $T = 0.60 \times 363.15\ \text{K} + 0.40 \times 288.15\ \text{K} = 333.15\ \text{K}$, which is $60.0\ ^\circ\text{C}$.
2. Hot water. $0.60\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1} \times \ln\dfrac{333.15\ \text{K}}{363.15\ \text{K}} = -216.6\ \text{J/K}$.
3. Cold water. $0.40\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1} \times \ln\dfrac{333.15\ \text{K}}{288.15\ \text{K}} = +243.0\ \text{J/K}$.
4. Total. $-216.6\ \text{J/K} + 243.0\ \text{J/K} = +26.4\ \text{J/K}$. It rises.

Practise with new numbers

Try it in Python

The program works out the mixture of the example.

Remember these

• $\Delta S = \dfrac{Q}{T}$ for heat taken in slowly at a steady temperature.
• Gas at a steady temperature: $\Delta S = nR\ln\dfrac{V_2}{V_1}$. Warming: $\Delta S = mc\ln\dfrac{T_2}{T_1}$.
• The total entropy of an isolated system never falls.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. $5000\ \text{J}$ of heat flows slowly into a large lake at $290\ \text{K}$. By how much does the lake's entropy rise?

2. $2.0$ mol of gas rushes into a vacuum, growing to $3$ times its volume, with no heat in and no change of temperature. What is its change in entropy?

3. Why does a broken glass never jump back together?

4. A hot stone in a river loses $12\ \text{J/K}$ of entropy. Does this break the second law?