Question

The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT/dt=-k(T-A), where T is the water temperature, A is the room temperature, and k is a positive constant.
If the water cools from 90°C to 85°C in 1 minute at a room temperature of 30°C, how long (to the nearest minute) will it take the water to cool to 60°C?

3
4
5 <- my answer
8

Answer 1

dT/dt = -k(T-A)
dT/(T-A) = -k dt
ln(T-A) = -kt
T-A = e^(-kt)
T = A + e^(-kt)

Now just plug in your numbers to find k and then solve for when T=60

Answer 2

Oops.
ln(T-A) = -kt + ln(c)
T-A = c*e^(-kt)
T = A + c*e^(-kt)

T(0) = 90, so

30 + c = 90
c = 60

T(t) = 30 + 60e^(-kt)
T(1) = 85, so
30 + 60e^-k = 85
e^-k = 55/60
-k = ln(55/60)
k = 0.087

T(t) = 30 + 60e^(-.087t)
we want when T=60:
30 + 60e^(-.087t) = 60
e^(-.087t) = 1/2
-.087t = -ln2
t = ln2/.087 = 7.96

Looks like 8 to me. How did you arrive at 5?