Question 77

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

Newton cooling: $$T-80=(T_0-80)e^{-Kt}$$.

At t=0: T₀=160, so T-80=$$80e^{-Kt}.$$

At t=15: 120-80=80$$e^{-15K}.$$ 40=80$$e^{-15K}$$. $$e^{-15K}$$=1/2.

At t=45: T-80=80$$e^{-45K}$$=80$$(e^{-15K})^3$$=80(1/2)³=10.

T=90°F.

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