Differential Equations Question 16
Question 16 - 2024 (31 Jan Shift 2)
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 positive constant. If $T(15)=120^{\circ} F$, then $T(45)$ is equal to
(1) $85^{\circ} F$
(2) $95^{\circ} F$
(3) $90^{\circ} F$
(4) $80^{\circ} F$
Show Answer
Answer (3)
Solution
$\frac{d T}{dt}=-k(T-80)$
$\int _{160}^{T} \frac{dT}{(T-80)}=\int _0^{t}-Kdt$
$[\ln |T-80|] _{160}^{T}=-kt$
$\ln |T-80|-\ln 80=-kt$
$\ln \left|\frac{T-80}{80}\right|=-kt$
$T=80+80 e^{-kt}$
$120=80+80 e^{-k \cdot 15}$
$\frac{40}{80}=e^{-k 15}=\frac{1}{2}$
$\therefore T(45)=80+80 e^{-k \cdot 45}$
$=80+80\left(e^{-k \cdot 15}\right)^{3}$
$=80+80 \times \frac{1}{8}$
$=90$
