Differential Equations Question 16
Question 16 - 2024 (31 Jan Shift 2)
The temperature $\mathrm{T}(\mathrm{t})$ of a body at time $\mathrm{t}=0$ is $160^{\circ} \mathrm{F}$ and it decreases continuously as per the differential equation $\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{K}(\mathrm{T}-80)$, where $\mathrm{K}$ is positive constant. If $\mathrm{T}(15)=120^{\circ} \mathrm{F}$, then $\mathrm{T}(45)$ is equal to
(1) $85^{\circ} \mathrm{F}$
(2) $95^{\circ} \mathrm{F}$
(3) $90^{\circ} \mathrm{F}$
(4) $80^{\circ} \mathrm{F}$
Show Answer
Answer (3)
Solution
$\frac{\mathrm{d} T}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80)$
$\int_{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_{0}^{\mathrm{t}}-\mathrm{Kdt}$
$[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}$
$\ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt}$
$\ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt}$
$\mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}}$
$120=80+80 \mathrm{e}^{-\mathrm{k} \cdot 15}$
$\frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2}$
$\therefore \mathrm{T}(45)=80+80 \mathrm{e}^{-\mathrm{k} \cdot 45}$
$=80+80\left(\mathrm{e}^{-\mathrm{k} \cdot 15}\right)^{3}$
$=80+80 \times \frac{1}{8}$
$=90$
