Newton's Law of Cooling - Pedagogy Zone

Statement

Newton’s law of cooling states that the rate of cooling of a body is directly proportional to the temperature difference between the body and the surroundings.

The law holds good only for a small difference of temperature. Loss of heat by radiation depends on the nature of the surface and the area of the exposed surface.

If θ be the temperature at the instant of the cooling body and $θ 0$ be the temperature of surroundings, the rate of loss of heat of the body

$\begin{aligned} \frac{\mathrm{dH}}{\mathrm{dt}} \propto\left(\theta-\theta_{0}\right) \\ -\frac{\mathrm{dH}}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta_{0}\right) \end{aligned}$ eqn (1)

Negative sign indicates that the heat is lost. The proportionality constant depends upon the shape, area and nature of the exposed surface.

Derivation

Consider a body of mass $m$ , specific heat $C$ and at temperature $θ$ . Let θ₀ be the temperature of the surroundings. Suppose, that temperature falls by a small amount $dθ$ in time $dt$ . Then the amount of heat lost.

dH = mCdθ

∴ Rate of loss of heat

eqn (2)

$\frac{\mathrm{dH}}{\mathrm{dt}}=\mathrm{mC} \frac{\mathrm{d} \theta}{\mathrm{dt}}$

From Newton’s law of cooling, the rate of heat is proportional to the difference of temperature (θ-θ₀)

eqn (3)

$-\frac{\mathrm{dH}}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta_{0}\right)$

Where K is a constant depending upon the area and the nature of the surface of the body.

From eqn. (2) and (3)

$\begin{aligned} -\mathrm{mC} \frac{\mathrm{d} \theta}{\mathrm{dt}} & =\mathrm{k}\left(\theta-\theta_{0}\right) \\ \frac{\mathrm{d} \theta}{\theta-\theta_{0}} & =-\frac{\mathrm{k}}{\mathrm{mC}} \cdot \mathrm{dt} \\ \frac{\mathrm{d} \theta}{\theta-\theta_{0}} & =-\mathrm{K} \cdot \mathrm{dt} \end{aligned}$ eqn (4)

K=(k/mC) being another constant and is independent of $t$ . Integrating both sides of eqn.(4)

$\begin{aligned} \int \frac{d \theta}{\theta-\theta_{0}} & =\int-\mathrm{K} \cdot \mathrm{dt} \\ \log _{\mathrm{e}}\left(\theta-\theta_{0}\right) & =-\mathrm{Kt}+\mathrm{c} \end{aligned}$