Question

PLEASE HELP WITH WORD PROBLEM(Newton's Law of Cooling)???
A cup of coffee was made at a temperature of 90C and cools according to Newton's law of cooling. The room temperature is 30C. If the temperature of the coffee 20 minutes after being made was 40 C
When was the temperature of the coffee 80C Time = how many minutes??

Answer 1

Hardly a word problem!

dT/dt = cooling = - k (T - Ta), where Ta is ambient (environment) temperature, 30 oC.

To = initial T = 90 oC

[(T - Ta) / (To - Ta)] = exp(-kt)

Take ln of both sides ln[ ] = - k t [natural logarithm, base e]

Solve for k from T=40 at t = 20 find that k = 0.020

Solve for t at T=80

ln[(80-30)/(90-30)] = ln[5/6] = -0.182 = - 0.020 t

t = 9 minutes