The temperature T(t) of a body at time t=0 is | vedclass

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(C) Given the differential equation: $\frac{dT}{dt} = -K(T-80)$.Separate the variables and integrate: $\int_{160}^{T} \frac{dT}{T-80} = \int_{0}^{t} -

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$90$

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(C) Given the differential equation: $\frac{dT}{dt} = -K(T-80)$.Separate the variables and integrate: $\int_{160}^{T} \frac{dT}{T-80} = \int_{0}^{t} -K dt$.This gives: $[\ln |T-80|]_{160}^{T} = -Kt$.$\ln |T-80| - \ln 80 = -Kt$.$\ln \left| \frac{T-80}{80} ight| = -Kt$,which implies $T-80 = 80e^{-Kt}$,or $T(t) = 80 + 80e^{-Kt}$.Given $T(15) = 120$,we have $120 = 80 + 80e^{-15K}$.$40 = 80e^{-15K} \implies e^{-15K} = \frac{40}{80} = \frac{1}{2}$.We need to find $T(45) = 80 + 80e^{-45K}$.$T(45) = 80 + 80(e^{-15K})^3$.Substituting $e^{-15K} = \frac{1}{2}$: $T(45) = 80 + 80 	imes (\frac{1}{2})^3$.$T(45) = 80 + 80 	imes \frac{1}{8} = 80 + 10 = 90^{\circ} F$.

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 a positive constant. If $T(15)=120^{\circ} F$,then $T(45)$ is equal to . . . . . . . . (in $^{\circ} F$)

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