Question

The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus 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?

Answer 1

dT/dt = -kT-A
dT/(kT+A) = -k dt
1/k ln(kT+A) = -kt + C
T(0) = 90 so 1/k ln(90k+30) = C
T(1) = 85 so 1/k ln(85k+30) = C-k

Solve for C and k, then evaluate t when

1/k ln(60k+30) = C-kt

Answer 2

dT/dt = -k (T-A) yes

try
T = A + (To - A) e^-kt
then
dT/dt = -k(To-A) e^-kt
but T-A = (To-A)e^-kt
so dT/dt = -k(T-A) it works

so
85 = 30 + (90-30)e^-k(1)
55 = 60 e^-k
ln(55/60) = -k solve for k

then use that k
60 = 30 +60 e^-k t
.5 = e^-kt
ln .5 = -k t solve for t

Answer 3

dT/dt = -kT-A
I think not
dT/dt = -k(T-A)