If the time taken by a body to cool from $50^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is 10 mi

Question

If the time taken by a body to cool from $50^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is 10 minutes when the surrounding temperature is $25^{\circ} \mathrm{C}$, then the taken for it to cool from $40^{\circ}$ of $30^{\circ} \mathrm{C}$ when the surrounding temperature is $15^{\circ} \mathrm{C}$, is\

Professor PC Thomas Classes · Answer

According to Newton's law of cooling\ $\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{t}_{1}}=\mathrm{k}\left[\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}ight)-\mathrm{T}_{0}ight]$\ When body cool from $50^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$\ $\left(\frac{50-40}{10}ight)=\mathrm{k}\left[\left(\frac{50+40}{2}ight)-25ight]$ $$\begin{equation*} 1=\mathrm{K}[45-25] \Rightarrow \mathrm{k}=\frac{1}{20} 	ag{i} \end{equation*}$$ Similarly when body cool room $40^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$, then\ $\frac{40-30}{\mathrm{t}_{2}}=\mathrm{k}\left[\frac{40+30}{2}-15ight]$ or $\frac{10}{\mathrm{t}_{2}}=\mathrm{k}[20]$ $$\begin{equation*} \Rightarrow \mathrm{k}=\frac{1}{2 \mathrm{t}_{2}} 	ag{ii} \end{equation*}$$ From equations (i) and (ii)\ $\frac{1}{20}=\frac{1}{2 \mathrm{t}_{2}} \Rightarrow \mathrm{t}_{2}=10 \mathrm{~min}$.\