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.

Question

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?
a) 3
b) 4
c) 5
d) 8 <--- my answer

Answer 1

dT/dt = -k(T-30)
dT/(T-30) = -k dt
ln(T-30) = -kt+C
T-30 = c*e^-k

T(0) = 90, so
c = 90-30 = 60
T-30 = 60e^-kt
T = 30+60e^-kt

T(1) = 85, so
30+60e^-k = 85
e^-k = 55/60
k = -ln 55/60 = 0.087
T = 30+60e^-.087t

30+60e^-.087t = 60
e^-.087t = 0.5
t = 7.967

So, I agree

Answer 2

yeahhhh. thank you