Newton’s law of cooling
How hot things cool
Pour a cup of hot chai and leave it on the table. At first it cools quickly. Later, as it nears room temperature, it cools more and more slowly.
Newton noticed that the rate of cooling depends on how much hotter the body is than its surroundings.
For a small temperature difference, the rate of loss of heat is proportional to the difference: minus dQ by dt equals k times T 2 minus T 1. T 2 is the body’s temperature, and T 1 that of the surroundings.
k is a positive constant that depends on the area and the nature of the body’s surface. Tea in a wide saucer cools faster than tea in a tall, narrow glass.
A body of mass m and specific heat s loses heat m s dT as it cools by dT. So the rate of fall of temperature is dT by dt equals minus K times T 2 minus T 1, with K equal to k over m s.
Rearrange and integrate. You get the natural log of T 2 minus T 1 equals minus K t, plus a constant.
So the excess temperature dies away exponentially. It starts at T 0 minus T 1, and is multiplied by e to the minus K t.
Watch the graph. It starts steep and flattens out towards room temperature, but never quite reaches it. The slope at any point is the rate of cooling.
Plot the log of T 2 minus T 1 against time instead, and you get a straight line with slope minus K. This is how we test the law.
Where does the law come from? Mostly from convection and radiation. For radiation, T to the fourth minus T s to the fourth is very nearly proportional to T minus T s, when the difference is small.
Testing and using the law
Here is the NCERT experiment. Hot water in a calorimeter sits inside a double-walled vessel, with water between the walls to keep the surroundings at a steady temperature. Read the thermometer every minute.
Plot temperature against time and you get the falling curve. Plot the log of T 2 minus T 1 against time, and it is a straight line, just as Newton’s law predicts.
Now a worked example. A pan of hot food cools from 94 to 86 degrees Celsius in 2 minutes, in a room at 20 degrees. How long will it take to cool from 71 to 69 degrees?
For a small drop, use the average temperature. From 94 to 86, the average is 90, which is 70 degrees above the room. The food cools 8 degrees in 2 minutes: 4 degrees per minute.
So 4 equals K times 70.
From 71 to 69, the average is 70, just 50 degrees above the room. The rate is K times 50, which is 4 times 50 over 70, about 2.86 degrees per minute.
To fall by 2 degrees takes 2 divided by 2.86, about 0.7 minutes, or 42 seconds.
Notice the idea: the excess temperature fell from 70 to 50, so the rate of cooling fell in the same ratio.
Remember that Newton’s law is an approximation. It works well when the temperature difference is small and the surroundings stay at a fixed temperature.
Try it: a cooling cup of tea
- Press Start the clock (or Space), or drag on the graph to move in time.
- The green tangent is the cooling rate. It shrinks as T approaches the room temperature.
- Tick “Plot ln(T − Tₛ)” to see a straight line of slope −k.
- The excess temperature halves every ln 2 / k minutes, whatever it started at.
Which graph is straight?
A body cools according to Newton’s law in surroundings at a fixed temperature Tₛ. Which graph is a straight line?
- T against t
- ln(T − Tₛ) against t
- T against ln t
- (T − Tₛ) against 1/t
Answer
B. Newton’s law gives dT/dt = −K(T − Tₛ). Integrating, ln(T − Tₛ) = −Kt + constant, a straight line of slope −K. The T–t graph is a curve that flattens out, because the cooling slows as the excess temperature shrinks.
Halving the excess
Tea at 90 °C is left in a room at 30 °C. It cools to 60 °C in 10 minutes. If it follows Newton’s law exactly, how much longer does it take to reach 45 °C?
- 5 minutes
- 10 minutes
- 15 minutes
- 20 minutes
Answer
B. The excess temperature T − Tₛ falls from 60 °C to 30 °C in 10 minutes: it halves. With T − Tₛ = (T₀ − Tₛ)e−Kt, it halves in every equal time interval. From 60 °C to 45 °C the excess falls from 30 to 15, another halving, so it takes another 10 minutes. Assuming a steady 3 °C per minute gives 5 minutes, but the cooling slows down.
Summary
- Newton’s law of cooling:
−dQ/dt = k(T₂ − T₁) - Rate of fall of temperature:
dT₂/dt = −K(T₂ − T₁), K = k/(ms) - Solution:
T − Tₛ = (T₀ − Tₛ) e^(−Kt) - Straight-line test:
ln(T − Tₛ) = −Kt + c
- The rate of cooling is proportional to the excess temperature over the surroundings, so hot bodies cool fast at first and slowly later.
- k depends on the surface area and nature of the surface; K = k/ms also depends on the mass and specific heat.
- For small temperature drops, use the average temperature of the interval to find the rate.
- The law holds only for small temperature differences and fixed surroundings.