A body at an unknow temperature is placed in a room which is held at a constant temperature of \(30^{\circ} \mathrm{F}\). If after 10 minutes the temperature of the body is \(0^{\circ} \mathrm{F}\) and after 20 minutes the temperature of the body is \(15^{\circ} \mathrm{F}\), then the expression for the temperature of the body at any time \(t\) is

We have \(\frac{\mathrm{dT}}{\mathrm{dt}} \propto(30-\mathrm{T})\)
\(
\begin{aligned}
& \therefore \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{K}(30-\mathrm{T}) \Rightarrow \int \frac{\mathrm{dT}}{30-\mathrm{T}}=\int-\mathrm{Kt} \\
& \therefore \log |30-\mathrm{T}|=-\mathrm{kt}+\mathrm{c}
\end{aligned}
\)
From given data, we write
\(
\begin{aligned}
& \log |30-0|=-10 K+c \\
& \log |30-15|=-20 K+c
\end{aligned}
\)
Solving (2) and (3), we get
\(
\log \left(\frac{30}{15}\right)=10 \mathrm{~K} \Rightarrow \mathrm{K}=\frac{1}{10} \log 2
\)
Substituting value of \(\mathrm{K}\) in eq. (2), we get
\(
\log 30=(-10)\left(\frac{\log 2}{10}\right)+\mathrm{c} \Rightarrow \mathrm{c}=\log 60
\)
Thus eq. (1) becomes
\(
\begin{aligned}
& \log |30-\mathrm{T}|=\frac{-\log 2}{10} \mathrm{t}+\log 60 \\
& \therefore \log \left|\frac{30-\mathrm{T}}{60}\right|=\frac{-0.3010}{10} \mathrm{t}=-0.03010 \mathrm{t} \\
& \therefore \frac{30-\mathrm{T}}{60}=\mathrm{e}^{-0.03010 \mathrm{t}} \quad \therefore \mathrm{T}=-60 \mathrm{e}^{-0.03010 \mathrm{t}}+30
\end{aligned}
\)