thank you! A cup of coffee at a temperture of 190 degree F is brought into a roo
ID: 2986765 • Letter: T
Question
thank you!
A cup of coffee at a temperture of 190 degree F is brought into a room. The temperature of the room is 70 degree F. After 10 min the temperature of the coffee is 170 degree F. Recall that the ditferential equation for newton's law of cooling is dT/dt = k (T-Tm). Start by writing down the solution of this equation (you don't need to show the details of solving the equation). Find the value of k for this problem. Clive an exact answer, not an approximation. Find the temperature of the coffee after 20 minutes. (give an exact answer and an approximation to two decimal places. At what time will the temperature of the coffee be 71 degree s? Give an exact answer and an approximation to two decimal places.Explanation / Answer
initial temperature of coffee=T0=190
tempertaure of room =Tm=70 F
temperature of coffee at time t=T(t)
y=T(t)-Tm
y0=T(0)-Tm
dy/dt=dT/dt-dTm/dt=dT/dt=-K(T-Tm)=-ky
dy/dt=-ky
solve above
y(t)=y0.e^-kt
y=T(t)-Tm
y0=T(0)-Tm
T(t)-T(m)=T0-Tm[e^-kt]
1.
Tt=70+(190-75)e^-kt
Tt=70+115e^-kt
after ten min T(t)=170 F
170=70+115e^-10k
solving above
-0.139767=-10k
k=0.013976 per min
2.
at 20 min
Tt=70+115e^-0.0139762*20
T20=156.956 F
3.
71=70+115e^-0.013976*t
t=339.5057 min
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