Newton’s law of cooling says that the rate at which a body cools is proportional to the difference C in temperature between the body and the environment around it

Question

The temperature f(t) of the body at time t in hours after being introduced to an environment having constant temperature T₀ is f(t)=T₀+Ce^-kt where C and k are constants.

A cup of coffee with temperatures 125 degree F is placed in a freezer with temperature 0 degree F. After 15 minutes, the temperature of the coffee is 37 degree F. Use Newton’s law of cooling to find the coffee’s temperature after 20 minutes.

Answer 1

f(t) = T₀ + Ce^-kt = temperature of the coffee t minutes after it is placed in the freezer

Given: T₀ = 0, and f(15) = 37. Find f(20)

f(t) = Ce^-15k

f(0) = Ce⁰ = 125. So, C = 125

f(t) = 125e^-kt

f(15) = 125e^-15k = 37

e^-15k = 0.296

-15k = ln(0.296)

k = 0.08116

So, f(t) = 125e^-0.08116t

f(20) = 125e^-0.08116(20) = 24.7°