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

Answer:

67.3°C

Step-by-step 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

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

$T(t) = 25 + 75e^{-kt}\\T(1) = 25 + 75e^{-k} = 90\\75e^{-k} = 65\\e^{-k} = \frac{65}{75}, k = ln\frac{75}{65} = 0.1431\\T(t) = 25+75e^{-0.1431t}\\T(4) = 25+75e^{-0.1431\times4} = 67.3$