Question

A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will it be 100◦F?

Answer 1

350 F to 100 F it take approx 87.33 min

Step-by-step explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling

`(dT)/(dt)` = -k(T-Ta)

`(dy)/(dt)` =
`(d)/(dt)` (T(t) -Ta)

=
`(dT)/(dt) -(dTa)/(dt) =(dT)/(dt)` = -k(T-Ta)

-ky
`(dy)/(dt)` = -ky

T(t) -Ta = (To -Ta)
`e^(-kt)` T(t) = Ta+ (To -Ta)
`e^(-kt)`

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 )
`e^(-k30)`

k = 0.025575

so here for T(t) = 100F

100 = 70 + ( 350 - 70 )
`e^(-0.025575*t)`

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min