1 Case study: Newton’s law of cooling

When a hot liquid is placed in a cooler environment, experimental observation shows that its temperature decreases to approximately that of its surroundings. A typical graph of the temperature of the liquid plotted against time is shown in Figure 1.

Figure 1

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After an initially rapid decrease the temperature changes progressively less rapidly and eventually the curve appears to ‘flatten out’.

Newton’s law of cooling states that the rate of cooling of liquid is proportional to the difference between its temperature and the temperature of its environment (the ambient temperature). To convert this into mathematics, let t be the time elapsed (in seconds, s), θ the temperature of the liquid (   ∘ C ),and θ 0 the temperature of the liquid at the start ( t = 0 ) . The temperature of the surroundings is denoted by θ s .

Task!

Write down the mathematical equation which is equivalent to Newton’s law of cooling and state the accompanying condition.

First, find an expression for the rate of cooling, and an expression for the difference between the liquid’s temperature and that of the environment:

The rate of cooling is the rate of change of temperature with time: d θ d t .

The temperature difference is θ − θ s . Now formulate Newton’s law of cooling:

You should obtain d θ d t r o p t o ( θ − θ s ) or, equivalently: d θ d t = − k ( θ − θ s ) . k is a positive constant of proportion and the negative sign is present because ( θ − θ s ) is positive, whereas d θ d t must be negative, since θ decreases with time. The units of k are s − 1 . The accompanying condition is θ = θ 0 at t = 0 which simply states the temperature of the liquid when the cooling begins.

In the above Task we call t the independent variable and θ the dependent variable. Since the condition is given at t = 0 we refer to it as an initial condition. For future reference, the solution of the above differential equation which satisfies the initial condition is θ = θ s + ( θ 0 − θ s ) e − k t .