Question

According to Newton's Law of Cooling

A cup of coffee with temperature of 130F is placed in a freezer with temperature 0F. After 5 minutes, the temperature of the coffee is 87F. Find the coffee's temperature

after 10 minutes.

Answer 1

Answer:- T(10) = 58.237

According to Newton's Cooling Law

T(t) = T_a+(T_o-T_a)e^-kt

where:

t is Time

T_a is the (constant) ambient temperature

T_o is the initial temperature of the object

k is a proportionality constant

Given:-  

T(5 Minutes) = 87

T_a = 0F

T_o = 130F

T(5) = 87 = 0 + (130 - 0 )e^-k(5)

I will solve this for k (in the units I have chosen):

87 = 0 + 130e^-5k

0.6692 = e^-5k

ln(0.6692) = lne^-5k

-0.40167 = -5k

k = 0.08033

Then at t = 10 minutes

T(t) = 130e^-(0.08033)t

T(10) = 130e^-(0.08033)10

^{T(10) = 58.237F}