Question

A hot body placed in a surrounding of temperature ${\theta }_0$ obeys newton's law of cooling $\frac{d\theta }{dt}=k(\theta -{\theta }_0)$ its temperature at t= 0 is ${\theta }_1$ . the specific heat capacity of the body s and its mass is m. find (a) the maximum heat that the body can lose and (b) the time starting from $t=0$ in which it will lose 90% of this maximum heat.

Answer 1

According to the newtons law of cooling,

$\frac{d\theta }{dt}=-K(\theta -{\theta }_0)$
temp at $t=0$ is ${\theta }_1$
(a) max. heat that the body can lose =$△Q_{max}=ms({\theta }_1-{\theta }_0)$
$\therefore as,△t=({\theta }_1-{\theta }_0)$
(b) if the body loses 90% of the max heat the decrease in its temp .will be
$\frac{△Q_m\times 9}{10ms}=\frac{({\theta }_1-{\theta }_0)\times 9}{10}$
if it makes time $t_1$ for this process the temp at $t_1$
${\theta }_1=({\theta }_1-{\theta }_0)\frac{9}{10}=\frac{10{\theta }_1-9{\theta }_1-9{\theta }_0}{10}=\frac{{\theta }_1-9{\theta }_0}{10}........\left(1\right)$
Now, $\frac{d\theta }{dt}=-K(\theta -{\theta }_1)$
Let $\theta ={\theta }_1$ at $t=0$ & $\theta$ be temp at time $t$
${\int }_{\theta }^{\theta }\frac{d\theta }{\theta -{\theta }_1}=-K{\int }_0^{t}dt$
or $ln\frac{{\theta }_1-\theta }{{\theta }_1-{\theta }_0}=-Kt$
or ${\theta }_1-\theta =({\theta }_1-{\theta }_0)e^{-kt}......\left(2\right)$
putting value in eq (1) and (2)
$\frac{{\theta }_{1}9{\theta }_0}{10}-{\theta }_0=({\theta }_1-{\theta }_0)e^{-kt}$
$t_1=\frac{ln10}{k}$