Question

The temperature of a cup of coffee varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.

74
67
60
42

Answer 1

67.3°C

Explanation:

A = 25, T(0) = 100, T(1) = 90, find T(4)

dT/dt = -k(T - A)

d(T - A)/dt = -k(T - A)

d(T - A)/(T - A) = -k dt

ln(T - A) = lnC - kt

`T(t) -A=Ce^(-kt)\\T(t) = A + Ce^(-kt) = 25 + Ce^(-kt)\\T(0) = 25 + C = 100\\C=75`

`T(t) = 25 + 75e^(-kt)\\T(1) = 25 + 75e^(-k) = 90\\75e^(-k) = 65\\e^(-k) = (65)/(75), k = ln(75)/(65) = 0.1431\\T(t) = 25+75e^(-0.1431t)\\T(4) = 25+75e^(-0.1431*4) = 67.3`