Differential Equations
Newton's Law of Cooling
nta_pyq_2024_jan
Grade 12
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 $\dfrac{dT}{dt}=-K(T-80)$, where $K$ is positive constant. If $T(15)=120^\circ F$, then $T(45)$ is equal to
$85^\circ F$
$95^\circ F$
$90^\circ F$
$80^\circ F$
Step-by-Step Solution
Key Concept: Solve $\frac{dT}{dt}=-K(T-80)$: $T-80=(T_0-80)e^{-Kt}=80e^{-Kt}$. Use $T(15)=120$ to find $e^{-15K}=\frac{1}{2}$. Then compute $T(45)$.
$T=80+80e^{-Kt}$. $e^{-15K}=\frac{1}{2}$. $T(45)=80+80\cdot\left(\frac{1}{2}\right)^3=80+10=90^\circ F$.
Correct Answer: 3