# Entropy and temperature

Add a little energy to a body at fixed volume and its entropy rises by that energy divided by its absolute temperature. The same energy raises the entropy of a cold body more than that of a hot one.

Written for this edition, not translated from Einstein.

## What does temperature have to do with entropy?

Add a small amount of heat to a body, slowly and reversibly, and keep its volume fixed so that it does no work. Its energy rises by that amount, and its entropy rises by the energy divided by its absolute temperature. So one joule raises the entropy of a body at 100 kelvin by 0.01 joule per kelvin, and of a body at 1000 kelvin by only 0.001: the same energy counts for more in a cold body. In symbols, with dE a small gain of energy and T the temperature:

$$
dS=\frac{dE}{T},\qquad\left(\frac{\partial S}{\partial E}\right)_V=\frac{1}{T}
$$

At fixed volume and the stated constraints, entropy change is energy change divided by absolute temperature.

Read the other way round, 1/T is how steeply a body's entropy climbs as its energy grows.

The subscript V matters. If the volume can change, the body can do work while it takes in energy, and the relation gains another term. The derivative describes one kind of change, at fixed volume, not every process.

The relation also says why bodies in contact end at one temperature. Share a fixed total of energy between two bodies, and the total entropy is greatest when moving a little energy from one to the other no longer changes it: when their entropy slopes, and so their temperatures, are equal. §3 of the light-quanta paper uses the same relation for radiation, setting its entropy slope equal to 1/T.

[Open the foundation: Partial derivatives and held-fixed quantities](/foundations/partial-derivatives/)

## Worked example: Three joules into a reservoir held at 300 kelvin

1. A reservoir so large that its temperature stays at 300 kelvin receives 3 joules of heat, slowly.
2. Its entropy rises by 3 J ÷ 300 K = 0.01 joule per kelvin.
3. The same 3 joules given to a reservoir at 150 kelvin would raise its entropy by 0.02 joule per kelvin, twice as much.
4. A small body warms as it takes in heat, so its temperature changes on the way. Its entropy change is found by adding dE/T over the warming, not by dividing by one temperature.

The relation fixes how entropy changes, not its absolute value: two entropy formulas that differ by a constant have the same slope everywhere, and a separate condition is needed to fix the constant. For an entropy per unit volume, that constant multiplied by two different volumes can change the difference between two states.

## Where this lesson stops

Before using 1/T as an entropy slope, say what is held fixed. Before comparing two entropies, say whether they are totals or per unit volume, and what fixes the constant.

## This lesson builds on

- [Partial derivatives and held-fixed quantities](/foundations/partial-derivatives/)
- [Adding continuously](/foundations/integration/)