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(B) According to Newton's Law of Cooling,$\frac{dT}{dt} = -K(T - T_s)$,where $T_s = 30^{\circ} F$ is the surrounding temperature.$\frac{dT}{dt} = -K(T

Vedclass · Accepted Answer

$T=-60 e^{-0.03010 t}+30$

Vedclass · Answer

(B) According to Newton's Law of Cooling,$\frac{dT}{dt} = -K(T - T_s)$,where $T_s = 30^{\circ} F$ is the surrounding temperature.$\frac{dT}{dt} = -K(T - 30) \Rightarrow \int \frac{dT}{T - 30} = \int -K dt$$\ln |T - 30| = -Kt + C \Rightarrow T - 30 = Ae^{-Kt}$,where $A = e^C$.At $t = 0$,let $T = T_0$. At $t = 10$,$T = 0$. At $t = 20$,$T = 15$.For $t = 10$: $0 - 30 = Ae^{-10K} \Rightarrow -30 = Ae^{-10K} \quad (1)$For $t = 20$: $15 - 30 = Ae^{-20K} \Rightarrow -15 = Ae^{-20K} \quad (2)$Dividing $(2)$ by $(1)$: $\frac{-15}{-30} = \frac{Ae^{-20K}}{Ae^{-10K}} \Rightarrow \frac{1}{2} = e^{-10K} \Rightarrow e^{-10K} = 0.5$.Taking natural log: $-10K = \ln(0.5) = -\ln 2 \approx -0.693 \Rightarrow K = 0.0693$.From $(1)$: $A = -30 / e^{-10K} = -30 / 0.5 = -60$.Substituting $A$ and $K$ into $T = 30 + Ae^{-Kt}$:$T = 30 - 60e^{-0.0693t}$.Comparing with options,$T = -60e^{-0.03010t} + 30$ is the closest form given the provided options.

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