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thepatient
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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.
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.
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
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.
Homework Statement
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.
Homework 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
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.
