The temperature T(t) of a body at t=0 is 160^ F and decreases per dTdt=-K(T-80), K0.

Question

The temperature $T(t)$ of a body at $t=0$ is $160^\circ F$ and decreases per $\dfrac{dT}{dt}=-K(T-80)$, $K>0$. If $T(15)=120^\circ F$, then $T(45)$ is:

Accepted Answer

$T-80 = 80e^{-Kt}.$

At $t=15$: $40=80e^{-15K} \Rightarrow e^{-15K}=\dfrac12$. At $t=45=3	imes15$: $T-80=80\left(\dfrac12ight)^3=10 \Rightarrow T=90^\circ F$.

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