Question by : A cup of coffee at 170 degrees is poured into a mug and left in a room at 65 degrees. After 1 minutes, the cof?
A cup of coffee at 170 degrees is poured into a mug and left in a room at 65 degrees. After 1 minutes, the coffee is 130 degrees. Assume that the differential equation describing Newton’s Law of Cooling is (in this case) (dT/dt)= k(T-65).
What is the temperature of the coffee after 11 minutes?
After how many minutes will the coffee be 100 degrees?
Best answer:
Answer by hfshaw
Newton’s law of cooling states that:
dT(t)/dt = -k*(T(t) – Ts)
where k is a positive constant, T is the temperature at time t, and Ts is the temperature of the surroundings. We can separate this differential equation to get:
dT/(T – Ts) = -k dt
If Ts is assumed to be constant, we can integrate to get:
ln(T – Ts) – ln(c) = -k*t
where ln(c) is the constant of integration.
If the temperature at t = 0 is To, then:
ln((To-Ts) – ln(c) = 0
c = (To – Ts)
So:
ln((T – Ts)/(To – Ts)) = -k*t
At this point, we can use the fact that at t = 1 min, the coffee has cooled from To = 170 to T(1min) = 130 in order to solve for the time constant “k”:
ln((130 – 65)/(170 – 65)) = -k*(1min)
ln(65/105) = -0.48 = -k*min
k = 0.48/min
Now solve the above equation for T:
T(t) = Ts + (To – Ts)*exp(-k*t)
Plugging in the values of the constants for this question:
T(t) = [65 + 105*exp(-0.48*t/min)]*degrees
Plug in t = 11min to find that after that amount of time, the coffee will be 65.5 degrees.
TO solve for the time at which the coffee will be a certain temperature, it’s easier to go back to the equation:
ln((T – Ts)/(To – Ts)) = -k*t
Solving for when the coffee is 100 degrees:
-(1min/0.48*ln((100 – 65)/(170 – 65)) = t(T)
t(100 deg) = 2.29 min
Know better? Leave your own answer in the comments!