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Question: Answered & Verified by Expert
A differential equation for the temperature ' $\mathrm{T}$ ' of a hot body as a function of time, when it is placed in a both which is held at a constant temperature of $32^{\circ} \mathrm{F}$, is given by (where $\mathrm{k}$ is a constant of proportionality)
MathematicsDifferential EquationsMHT CETMHT CET 2021 (20 Sep Shift 1)
Options:
  • A $\frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{kT}-32$
  • B $\frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{kT}+32$
  • C $\frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{k}(\mathrm{T}-32)$
  • D $\frac{\mathrm{dT}}{\mathrm{dt}}=32 \mathrm{kT}$
Solution:
1671 Upvotes Verified Answer
The correct answer is: $\frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{k}(\mathrm{T}-32)$
The temperature $\mathrm{T}$ of the body will decrease with time. The body is kept in a bath of temperature $32^{\circ} \mathrm{F}$.
$$
\begin{aligned}
& \therefore \frac{\mathrm{dT}}{\mathrm{dt}} \alpha-(\mathrm{T}-32) \\
& \Rightarrow \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-32)
\end{aligned}
$$

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