Newton's law of cooling

Fast at first, then slowly

A cup of tea is too hot to drink, then just right for a while, and finally cold. It cools quickly while it is much hotter than the room, and more and more slowly as it nears the room's temperature.

Newton's law of cooling: the rate of cooling is in proportion to the excess temperature,

where $T_0$ is the surroundings' temperature and $k$ depends on the body's surface, area and heat capacity. The law holds well for modest differences; it follows from Stefan's law and also describes cooling by convection.

The cooling curve

The excess temperature falls by the same fraction in equal times:

It halves every $\dfrac{\ln 2}{k}$, whatever the starting temperature. The body never quite reaches, and never passes, the room's temperature.

You should find that halving $k$ doubles the time to fall halfway, and that the starting temperature does not change that time.

A quick estimate

Over a short interval, use the average excess. If a body cools by $10\ \text{K}$ in time $t_1$ with an average excess $\theta_1$, then the next $10\ \text{K}$, with average excess $\theta_2$, takes $t_1\dfrac{\theta_1}{\theta_2}$.

Watch out. It is the excess over the surroundings that decays, not the temperature itself.

A worked example

Tea cools from $90\ ^\circ\text{C}$ to $70\ ^\circ\text{C}$ in $6.0$ minutes in a room at $25\ ^\circ\text{C}$. When will it be at $40\ ^\circ\text{C}$?

1. The constant. The excess falls from $65\ \text{K}$ to $45\ \text{K}$: $k = \dfrac{1}{6.0\ \text{min}}\ln\dfrac{65\ \text{K}}{45\ \text{K}} = 0.0613\ \text{min}^{-1}$ per minute.
2. To $40\ ^\circ\text{C}$. The excess must fall to $15\ \text{K}$: $t = \dfrac{1}{0.0613\ \text{min}^{-1}}\ln\dfrac{65\ \text{K}}{15\ \text{K}} = 23.9\ \text{min}$ minutes.

Practise with new numbers

Try it in Python

The program finds $k$ for the tea, then steps Newton's law forward and compares with the formula.

Remember these

• $\dfrac{dT}{dt} = -k(T - T_0)$.
• $T = T_0 + (T_s - T_0)e^{-kt}$; the excess halves every $\dfrac{\ln 2}{k}$.
• Quick estimate: rate in proportion to the average excess.

Check yourself

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

1. A body $40\ \text{K}$ above the room cools at $2.0\ \text{K/min}$. How fast does it cool when $10\ \text{K}$ above?

2. Can a cup of tea cool below room temperature by itself?

3. The excess temperature of a body halves every $8.0$ minutes. What fraction of it is left after $24$ minutes?

4. Why does stirring hot soup make it cool faster?