What is the answer to the problem?

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 Cooling 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 that is associated with the cooling object. A cake is removed from the oven and has a temp of 222 Degrees F and is left to cool in a room that has a temp of 74 degrees F. After 40 min the temp of the cake is 148 degrees F. What is the temp of the cake after 50 min?

Answer 1

T = C + (t-C)e^-kt

in 40 minutes the cake goes from 222 to 148 degrees

with the room temperature = 74. what is the cake's temperature after another 10 minutes. How much less than 148 degrees?

148 = 74 + (222-74)e^-40k

148-74 = 148e^-40k

74 = 148e^-40k

1/2 = e^-40k

take natural logs

ln(.5) = -40k

k = ln.5/-40 = 0.01734

after 50 minutes

A -74 = 148e^-50(.01734) = 148e^-.86643 = 148(.4204) = 62.226

A = 74+ 62.226 = 136.226 degrees after 50 minutes

the cake's temperature = about 136 degrees F after 50 minutes

It went from 148 to 136, dropping 12 degrees in 10 minutes