# Newton's Law of cooling DE involving delicious coffee.

1. Feb 4, 2012

### thepatient

I did this problem a few times now and double checked my procedure but I don't see what I did wrong. The answer should be 6.07 minutes according to the back of the book, but I get 3.63.

1. The problem statement, all variables and given/known data
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 200 degrees F when poured, 1 minute later cooled to 190 degrees F, room temperature 70 deg F, determine time at which temperature is 150 degrees F.

2. Relevant equations
dT/dt = k(Ts - T(t))

k being constant of proportionality.
Ts ambient temperature
T(t) temperature of cup of coffee at a time t.

T(0) = 200
T(1) = 190

3. The attempt at a solution

dT/dt + kT(t) = kTs

multiplying equation by integration factor mu = e^(kt)

(T(t) * e^(kt))' = k*Ts*e^(kt)

Integrating both sides:

T(t) * e^(kt) = Ts*e^(kt) + c
T(t) = Ts + c*e^(-kt)

Since Ts = 70, and T(0) = 200:

200 = 70 + c
c = 140

T(t) = 70 + 140e^(-kt)

T(1) = 190:

190-70 = 140 e^(-k)
6/7 = e^(-k)

ln(7/6) = k

Therefore, equation for temperature of coffee at any time t:
T(t) = 70 + 140e^(ln(6/7)t)

Solving for T(t) = 150:

(150-70)/140 = e^(tln6/7)
ln(4/7) = tln(6/7)
t = ln(4/7)/ln(6/7) = 3.63 minutes.

Can anyone see if maybe I made a mistake? Thanks a ton.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 4, 2012

200-70=????

ehild

3. Feb 4, 2012

### thepatient

Hehehe. Looks like I need to go back to arithmetic class. XD Such a insignificant part of the calculation it seemed that I overlooked it several times. Thanks a ton.

4. Feb 4, 2012

### ehild

Such things happen to everybody including myself.

ehild