Over a period of time, a hot object cools to the temperature of the surrounding air. This is described mathematically as Newtons Law of Cooing T=C+(t_0-C)e^-kt, where t is the time it takes for an object to cool from the temp T_0 to the temp T, C is the surrounding air temp, and k is a positive constant tat is associated with the cooling object. A cakes removed from the oven has a temp of 207 Degrees F and is left to cool in a room that has a temp of 69 degrees F. After 20 min the temp of the cake is 138 degrees F. What is the temp of the cake after 35 min?

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Tringa0 Tringa0 · Answer 1 · 2019-08-08T07:33:22+02:00

Answer:

110.03 °F is the temp of the cake after 35 minutes.

Explanation:

Newtons Law of Cooing is given by:

[tex]T=C+(T_o-C)e^{-kt}[/tex]

t = Time it taken by an object to cool from the temp [tex]T_o[/tex] to the temp T

C = temperature of the surrounding

k = Constant that is associated with the cooling object

We will first calculate the value of k.

[tex]T_o=207^oF,T=138^oF,C=69^oF,t=20 min[/tex]

Substituting the values in a given formula:

[tex]138^oF=69^oF+(207^oF-69^oF).e^{-k20 min}[/tex]

[tex]k=0.03465 min^{-1}[/tex]

The temperature of the cake after 35 min:

[tex]T_o=207^oF,T=?,C=69^oF,t=35 min,k=0.03465 min^{-1}[/tex]

[tex]T=69^oF+(207^oF-69^oF).e^{-(0.03465 min^{-1})35 min}[/tex]

T = 110.03 °F

110.03 °F is the temp of the cake after 35 minutes.