Newton's Law of Cooling

Question

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

A cup of coffee with temperature of 120 degrees Fahrenheit is placed in a freezer with temperature of 0 degrees Fahrenheit. After 10 minutes, the temperature of the coffee is 53 degrees Fahrenheit. Using Newton's Law of Cooling to find the coffee's temperature after 20 minutes.

Answer 1

Hi Radhika:

Let's do this one!

Newton's Law of Cooling: T(t) = T₀ + Ce^-kt

where T(t) = temp at time t , T₀ = initial temp (at t = 0), C = temp difference between the body and the environment

Apply this Law to the coffee 53 = 0 + (120 - 0)e^-k(10/60)

=> 53 = 120e^-k/6

=> e^-k/6 = 53/120 [reversing the equation for convenience

=> -k/6 = log_e (53/120)

=> -k/6 = -0.8171

=> k = 0.8171 x 6

= 4.9

∴ T(20) = 120 e^-k(20/60)

= 120 e^-4.9/3

= 23.4 degrees F