State and explain Newtons law of cooling state the conditions under which Newton's law of cooling is applicable. A body cools down from 60°C to 50°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surroundings.

Statement :

The rate of loss of heat of a hot body is directly proportional to the difference in temperature between the body and its surroundings providing the temperature difference is small.

Explanation:

According to Newtons law of cooling

$\frac{-\mathrm{dQ}}{\mathrm{dt}}\mathrm{\alpha }\left(\mathrm{\theta }-{\mathrm{\theta }}_0\right) (-\mathrm{vesignindicateslossofheat})$
where  $\mathrm{\theta }=$temperature of the hot body &

$\mathrm{\theta }=$temperature of the surroundings

$\therefore \frac{\mathrm{dQ}}{\mathrm{dt}}=-\mathrm{K}\left(\mathrm{\theta }-{\mathrm{\theta }}_0\right)$

Where ' k ' is a proportionality constant.

we knows that $\mathrm{dQ}=\mathrm{msd\theta }$

$\Rightarrow \frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{ms}\frac{\mathrm{d\theta }}{\mathrm{dt}}$.....(2)

From (1) and (2)
$\begin{array}{r}\mathrm{ms}\frac{\mathrm{d\theta }}{\mathrm{dt}}=-\mathrm{K}\left(\mathrm{\theta }-{\mathrm{\theta }}_0\right)\\ \Rightarrow \frac{\mathrm{d\theta }}{\mathrm{dt}}=\frac{-\mathrm{k}}{\mathrm{ms}}\left(\mathrm{\theta }-{\mathrm{\theta }}_0\right)\end{array}$

$\frac{\mathrm{k}}{\mathrm{ms}}$is constant independent of time.

Newtons law of cooling is applicable:

i) Loss of heat is negligible by conduction and it is only due to convection.

ii) Loss of heat occurs in a stream lined flow of air i.e forced convection.

iii)Temperature of the body is uniformly distributed over it.

iv)Temperature differences are moderate.
 

Problem:
According to Newtons law of cooling
$\frac{\mathrm{d\theta }}{\mathrm{dt}}=\mathrm{k}\left(\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}-{\mathrm{\theta }}_0\right)$

Case I :

$\begin{array}{l}{\mathrm{\theta }}_1={60}^{\circ },{\mathrm{\theta }}_2={50}^{\circ }\mathrm{C},\mathrm{t}=5\min =5\times 60\sec \\ \Rightarrow \frac{60-50}{5\times 60}=\mathrm{k}\left(\frac{60+50}{2}-{\mathrm{\theta }}_0\right)\\ \frac{10}{5\times 60}=\mathrm{K}(55-\mathrm{\theta })\text{.... (1) }\end{array}$

Case II :

$\begin{array}{l}{\mathrm{\theta }}_1={50}^{\circ }\mathrm{C},{\mathrm{\theta }}_2={40}^{\circ }\mathrm{C},\mathrm{t}=8\min =8\times 60\sec \\ \Rightarrow \frac{50-40}{8\times 60}=\mathrm{k}\left(\frac{50+40}{2}-{\mathrm{\theta }}_0\right)\\ \left(1\right)/\left(2\right)\frac{8}{5}=\frac{55-{\mathrm{\theta }}_0}{45-{\mathrm{\theta }}_0}\\ \Rightarrow 8\left(45-{\mathrm{\theta }}_0\right)=5\left(55-{\mathrm{\theta }}_0\right)\\ \Rightarrow 3{\mathrm{\theta }}_0=360-275=85\\ \Rightarrow {\mathrm{\theta }}_0=\frac{85}{3}=28{.33}^{\circ }\mathrm{C}\end{array}$