Newton’s Law of Cooling

NEWTON'S LAW OF COOLING

According to this law, if the temperature T of the body is not very different from that of the surroundings T₀, then rate of cooling –$\frac{dT}{dt}$ is proportional to the temperature difference between them. T₀ prove it let us assume that

T = T₀ + $\Delta$T

So that T⁴ = (T₀ + $\Delta$T)⁴ = $T_0^4{\left(1+\frac{\Delta T}{T_0}\right)}^4$

$\approx T_0^4\left(1+\frac{4\Delta T}{T_0}\right)$ (from binomial expansion)

$\therefore$ (T⁴ – $T_0^4$) = 4$T_0^3$ ($\Delta$T)

or (T⁴ – $T_0^4$) $\propto$ $\Delta$T (as T₀ = constant)

Now, we have already shown that rate of cooling

$\left(-\frac{dT}{dt}\right)\propto (T^4-T_0^4)$

and here we have shown that

$(T^4-T_0^4)\propto \Delta T$,

if the temperature difference is small.

Thus, rate of cooling

$\left(-\frac{dT}{dt}\right)\propto \Delta T$ or $-\frac{d\theta }{dt}\propto \Delta \theta$

as dT = d$\theta$ or $\Delta$T = $\Delta$$\theta$

Variation of temperature of a body according to Newton's law

Suppose a body has a temperature $\theta$_i at time t = 0. It is placed in an atmosphere whose temperature is $\theta$₀. We are interested in finding the temperature of the body at time t, assuming Newton's law of cooling to hold good or by assuming that the temperature difference is small. As per this law,

rate of cooling $\propto$ temperature difference

or $\left(-\frac{d\theta }{dt}\right)=\left(\frac{eA\sigma }{mc}\right)(4{\theta }_0^3)\left(\theta -{\theta }_0\right)$

or $\left(-\frac{d\theta }{dt}\right)=\alpha \left(\theta -{\theta }_0\right)$

Here a = $\left(\frac{4eA\sigma {\theta }_0^3}{mc}\right)$ is a constant

$\therefore$ ${\int }_{{\theta }_i}^{\theta }\frac{d\theta }{\theta -{\theta }_0}=-\alpha {\int }_0^{t}dt$

$\therefore$

$\theta ={\theta }_0+\left({\theta }_i-{\theta }_0\right)e^{-\alpha t}$

From this expression we see that $\theta$ = $\theta$_i at t = 0 and $\theta$ = $\theta$₀ at t = $\infty$, i.e. temperature of the body varies exponentially with time from $\theta$_i to $\theta$₀ (<$\theta$_i). The temperature versus time graph is as shown in Fig.

Note: If the body cools by radiation from $\theta$₁ to $\theta$₂ in time t, then taking the approximation

$\left(-\frac{d\theta }{dt}\right)=\frac{{\theta }_1-{\theta }_2}{t}$ and

$\theta$ = $\theta$_av = $\left(\frac{{\theta }_1-{\theta }_2}{2}\right)$

The equation $\left(-\frac{d\theta }{dt}\right)=\alpha \left(\theta -{\theta }_0\right)$ becomes

$\frac{{\theta }_1-{\theta }_2}{t}=\alpha \left(\frac{{\theta }_1-{\theta }_2}{2}-{\theta }_0\right)$

This form of the low helps in solving numerical problems related to Newton's law of cooling.