The rate at which a substance cools in moving | vedclass

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Question

(A) According to Newton's Law of Cooling,$\frac{dT}{dt} = -k(T - T_a)$,where $T_a = 290 \ K$. Integrating,we get $\ln(T - 290) = -kt + C$,or $T - 290

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) According to Newton's Law of Cooling,$\frac{dT}{dt} = -k(T - T_a)$,where $T_a = 290 \ K$. Integrating,we get $\ln(T - 290) = -kt + C$,or $T - 290 = Ce^{-kt}$. At $t = 0$,$T = 370$,so $370 - 290 = C$,which gives $C = 80$. Thus,$T - 290 = 80e^{-kt}$. At $t = 10 \ min$,$T = 330$,so $330 - 290 = 80e^{-10k}$,which means $40 = 80e^{-10k}$,so $e^{-10k} = 0.5$. Taking natural logs,$-10k = \ln(0.5) = -\ln(2)$,so $k = \frac{\ln(2)}{10}$. We want to find $t$ when $T = 295$. $295 - 290 = 80e^{-kt} \implies 5 = 80e^{-kt} \implies e^{-kt} = \frac{5}{80} = \frac{1}{16} = (\frac{1}{2})^4$. Since $e^{-10k} = \frac{1}{2}$,we have $e^{-kt} = (e^{-10k})^4 = e^{-40k}$. Therefore,$kt = 40k$,which gives $t = 40 \ minutes$.

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