A salad with a temperature of 43F is taken from the refrigerator and placed on the table in a room that is 68F. After 12 min the temperature of the salad is 55F. What will the temperature of the salad be after 20 min.

Answer 1

well, you know that

dT/dt = k(68-T)
dT/(68-T) = k dt
ln(68-T) = kt + c1
68-T = e^(kt+c1)
68-T = ce^(kt)
T = 68 - ce^(kt)

We know that T(0) = 43, so c=25, and

T = 68-45e^(kt)

We know that T(12) = 55, so

55 = 68-45e^(12k)
45e^12k = 13
e^12k = (13/45)
12k = ln(13/45)
k = ln(13/45)/12 = -0.103

So, we have

T(t) = 68 - 45e^(-0.103t)

Now just find T(20)