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A body at an unknown temperature is placed in a room which is held at a constant temperature of $30^{\circ}…
A body at an unknown 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
$\mathrm{T}=-60 \mathrm{e}^{-0.069 \mathrm{t}}-30$ $\mathrm{T}=-60 \mathrm{e}^{-0.03010 \mathrm{t}}+30$ $\mathrm{T}=60 \mathrm{e}^{-0.069 \mathrm{t}}+30$ $\mathrm{T}=60 \mathrm{e}^{-0.069 \mathrm{t}}-30$ Solution
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
$\begin{aligned}
& \log |30-\mathrm{T}|=\frac{-\log 2}{10} \mathrm{t}+\log 60 \\
& \therefore \log \left|\frac{30-\mathrm{T}}{60}\right|=\frac{-0.3010}{10} \mathrm{t}=-0.03010 \mathrm{t} \\
& \therefore \frac{30-\mathrm{T}}{60}=\mathrm{e}^{-0.03010 \mathrm{t}} \quad \therefore \mathrm{T}=-60 \mathrm{e}^{-0.03010 \mathrm{t}}+30
\end{aligned}$
Asked in: MHT CET 2021 (21 Sep Shift 1)
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