The temperature T(t) of a body at time t=0 is 160^ F and it decreases continuously as per …

Question

The temperature $T(t)$ of a body at time $t=0$ is ${160}^{\circ }F$ and it decreases continuously as per the differential equation $\frac{dT}{dt}=-K(T-80)$, where $K$ is a positive constant. If $T(15)={120}^{\circ }F$, then $T(45)$ is equal to

Accepted Answer

$$\begin{aligned}
& \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \
& \int_\limits{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_\limits0^{\mathrm{t}}-\mathrm{Kdt} \
& {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \
& \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt} \
& \ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt} \
& \mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}} \
& 120=80+80 \mathrm{e}^{-\mathrm{k} .15} \
& \frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2} \
& \therefore \mathrm{T}(45)=80+80 \mathrm{e}^{-\mathrm{k} .45} \
& =80+80\left(\mathrm{e}^{-\mathrm{k} .15}\right)^3 \
& =80+80 \times \frac{1}{8} \
& =90
\end{aligned}$$

Solutions · 1