Use the Excess Over Surroundings, Not the Absolute Temperature
Cooling problems are about the excess temperature over the surroundings. For small differences Newton's law of cooling applies; for large differences radiation follows Stefan's law instead. Choose the law the question describes and apply it consistently.
Cooling Rate and the Law of Cooling is the central idea in this question pattern. Identify the condition in the stem before applying the method below.
Newton's law and the average-temperature shortcut
Integrating, ln(T - Ts) falls linearly with time, so a graph of log(T - Ts) against t is a straight line with negative slope, and the excess over the surroundings falls by the same factor in equal times.
A body cooling from 80 °C to 60 °C in 10 minutes in surroundings at 30 °C gives 20/10 = k(70 - 30), so k = 0.05 per minute; for the next 10 minutes (60 - T)/10 = 0.05[(60 + T)/2 - 30], so T = 48 °C.
Using the same method for other data
A body that cools from 3T to 2T in 15 minutes in a room at T has its excess fall from 2T to T, halving in 15 minutes, so after another 15 minutes the excess is T/2 and the temperature is 1.5T; the average-temperature method gives the same 1.5T. Whenever the excess halves (or falls by any fixed factor) in an interval, you can chain intervals directly. The law is also the basis of the laboratory method for comparing the specific heat of a liquid with that of water by timing their cooling.
Cooling rate from radiation and the effect of size
When the cooling is by radiation, the rate of cooling is proportional to σ e A (T⁴ - Ts⁴) / (m s): at 600 K in a 200 K environment the rate scales as (600⁴ - 200⁴), and the same formula at 400 K scales as (400⁴ - 200⁴). For spheres of the same material at the same temperature, A is proportional to r² and the mass to r³ ρ, so the cooling rate is proportional to 1/(r ρ): spheres with radii in the ratio 1 : 2 and densities in the ratio 2 : 1 cool at the same rate. For equal surface areas a sphere holds more mass than a cube, so it cools more slowly.
OlympIQ exam lens
- ✓Newton: -dT/dt = k(T - Ts); the excess over the surroundings decays exponentially, so a log plot is a straight line
- ✓For a single interval, (T1 - T2)/t = k[(T1 + T2)/2 - Ts]
- ✓If the excess falls by a fixed factor in one interval it falls by the same factor in the next interval of equal length
- ✓Radiative cooling rate is proportional to σ e A (T⁴ - Ts⁴) / (m s); do not mix this with Newton's linear law
- ✓For equal surface area, the sphere has the most mass, so it cools most slowly; for the same material, rate is proportional to 1/(r ρ)
Do not use the body's absolute temperature, or the ratio of temperatures, in a cooling problem. Newton's law depends on the excess over the surroundings, so subtract the surrounding temperature first; and if the question describes radiation, use T⁴ - Ts⁴ in kelvin, not a linear difference.
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