Answer:
350 F to 100 F it take approx 87.33 min
Explanation:
given data
oven = 350◦F
cooling rack = 70◦F
time = 30 min
cake = 200◦F
solution
we apply here Newtons law of cooling
[tex]\frac{dT}{dt}[/tex] = -k(T-Ta)
[tex]\frac{dy}{dt}[/tex] = [tex]\frac{d}{dt}[/tex] (T(t) -Ta)
= [tex]\frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt}[/tex] = -k(T-Ta)
-ky [tex]\frac{dy}{dt}[/tex] = -ky
T(t) -Ta = (To -Ta) [tex]e^{-kt}[/tex] T(t) = Ta+ (To -Ta) [tex]e^{-kt}[/tex]
put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F
so here
200 = 70 + ( 350 - 70 ) [tex]e^{-k30}[/tex]
k = 0.025575
so here for T(t) = 100F
100 = 70 + ( 350 - 70 ) [tex]e^{-0.025575*t}[/tex]
time = 87.33 min
so here 350 F to 100 F it take approx 87.33 min
