Question

A hot body placed in a surrounding of temperature θ₀ obeys Newton's law of cooling $\frac{d\mathrm{\theta }}{dt}=-k\left(\mathrm{\theta }-{\mathrm{\theta }}_0\right)$. Its temperature at t = 0 is θ₁. 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

According to Newton's law of cooling,
$\frac{d\mathrm{\theta }}{dt}=-k\left(\mathrm{\theta }-{\mathrm{\theta }}_0\right)$

(a) Maximum heat that the body can lose, ∆Q_max = ms (θ₁ − θ₀)

(b) If the body loses 90% of the maximum heat, then the fall in temperature will be θ_.
$$\begin{gathered} ∆Q_{\max }\times \frac{90}{100}=ms({\mathrm{\theta }}_1-\mathrm{\theta }) \\ \Rightarrow \mathrm{m}\mathrm{s}({\mathrm{\theta }}_1-{\mathrm{\theta }}_0)\times \frac{9}{10}=\mathrm{m}\mathrm{s}({\mathrm{\theta }}_1-\mathrm{\theta }) \\ \Rightarrow \mathrm{\theta }={\mathrm{\theta }}_1-({\mathrm{\theta }}_1-{\mathrm{\theta }}_0)\times \frac{9}{10} \\ \Rightarrow \mathrm{\theta }=\frac{{\mathrm{\theta }}_1-9{\mathrm{\theta }}_0}{10}...\left(\mathrm{i}\right) \end{gathered}$$

From Newton's law of cooling,
$\frac{d\theta }{dt}=-k\left({\mathrm{\theta }}_1-\mathrm{\theta }\right)$
Integrating this equation within the proper limit, we get
At time t = 0,
θ = θ₁
At time t,
θ = θ

$$\begin{gathered} {\int }_{{\mathrm{\theta }}_1}^{\mathrm{\theta }}\frac{d\mathrm{\theta }}{{\mathrm{\theta }}_1-\mathrm{\theta }}=-k{\int }_0^{t}dt \\ \Rightarrow \ln \left(\frac{{\mathrm{\theta }}_1-\mathrm{\theta }}{{\mathrm{\theta }}_1-{\mathrm{\theta }}_0}\right)=-kt \\ \Rightarrow {\mathrm{\theta }}_1-\mathrm{\theta }={\mathrm{\theta }}_1-{\mathrm{\theta }}_0{e}^{-kt}...\left(\mathrm{ii}\right) \end{gathered}$$

From (i) and (ii),

$$\begin{gathered} \frac{{\mathrm{\theta }}_1-9{\mathrm{\theta }}_0}{10}-{\mathrm{\theta }}_0=\left({\mathrm{\theta }}_1-{\mathrm{\theta }}_0\right)e^{-kt} \\ \Rightarrow t=\frac{\ln (10}{k}) \end{gathered}$$