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 is 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

$\left(\frac{d\theta }{dt}\right)=-K(\theta -{\theta }_0)$
Temperature at

t= 0 is ${\theta }_1$

(a) Maximum heat that the body can loose

= $\Delta Qm=ms({\theta }_1-{\theta }_0)$

$(\because dt={\theta }_1-{\theta }_0)$

(b) If the body loses 90 % of the maximum heat the decrease in its temperature will be

$\frac{\Delta Q_m\times 9}{10ms}=\frac{({\theta }_1-{\theta }_0)\times 9}{10ms}$

If it takes time $t_1$, for this process, the temperature at $t_1$

= ${\theta }_1-({\theta }_1-{\theta }_0)\frac{9}{10}$

= $\frac{10{\theta }_1-9{\theta }_1-9{\theta }_0}{10}$

= $\frac{{\theta }_1-9{\theta }_0}{10}$ ...(i)

Now, $\left(\frac{d\theta }{dt}\right)=-K({\theta }_1-{\theta }_0)$

Let $\theta ={\theta }_1$ at t = 0; and $\Theta$ be the temperature at time $t_1$

${\int }_{{\theta }_1}^{\theta }\left(\frac{d\theta }{\theta -{\theta }_0}\right)=-K{\int }_t^0$ dt

or, $({\theta }_1-\theta )=({\theta }_1-{\theta }_0)e^{-kt}$

Putting value in the Equation (2) and Equation(1)

$\frac{{\theta }_1-9{\theta }_0}{10}-{\theta }_0=({\theta }_1-{\theta }_0)e^{-kt}$

$\Rightarrow t_1$ = ln $\frac{10}{k}$