Newton's Law of Cooling WP

Quote:

Originally Posted by pham07

When the Jolie family left the house yesterday morning, they turned off the heat to save energy. At the time they left the house, the indoor temperature was 70 degrees and the outdoor temperature was 30. In two hours, the house cooled down to 60 degrees.

Assuming that the outdoor temperature remains constant throughout the day, the house loses heat according to Newton's Law of Cooling.

Solve the IVP and determine how long it takes from the time the heat is turned off, for the house temperature to reach 50 degrees.

I got the first part. I found my model and found the time. What I don't get is the next part.

When the tempature reaches 50 degrees, the heat is turned back on, warming the house at a rate of 10 degrees per hour. Modify the DE to account for the heat being turned on.
Solve the modified DE and determine the house temperature will be 2 hours after the heat is turned on. The value of k is still valid.

Would the new DE become dT/dt= k(t-Tm)+10t ? where Tm is the outside temp

Hello pham07,

No the equation becomes:

$\frac{dT}{dt}=10-k(T-T_w)$

Solving this and applying the boundary condition at t=0 is T=50 gives:

$T=\left(50-T_w-\frac{10}{k}\right)e^{-kt}+T_w+\frac{10}{k}$

From this you can find the temperature at t=2.

best regards,

Coomast