Temperature

Question

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?

Answer 1

You're told that t₀ = 207 (initial temperature) and C = 69 (surrounding air temperature). You're also told that when t = 20 (minutes that have passed), T = 138 (current temperature). Plug those values into the equation and solve for k:

 

138 = 69 + (207 - 69)e^-k(20)

69 = 138e^-20k

1/2 = e^-20k

ln(1/2) = -20k

-0.6931 = -20k

0.03466 = k

 

 

Now that you know k = 0.03466, plug that in, along with t₀ = 207, C = 69, and t = 35 to solve for T (the temperature after 35 minutes):

 

T = 69 + (207 - 69)e^-0.03466(35)

T = 69 + 138e^-1.213

T = 69 + 138(0.2973)

T = 69 + 41.028

T = 110.03

 

The temperature is approximately 110 degrees F.