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=80e^{-15K}. 40=80e^{-15K}. e^{-15K}=1/2.

At t=45: T-80=80e^{-45K}=80(e^{-15K})³=80(1/2)³=10. T=90°F.

The answer is Option (3): 90°F.

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