Newton's Law of Cooling

Submitted by charles-d-nunez on

Hello, I need help with this question:

On a hot Saturday morning while people are working inside, the air conditioner keeps the temperature inside the building at 24⁰C. At noon, the air conditioner is turned off, and the people go home. The temperature outside is a constant 35⁰C for the rest of the afternoon. If the time constant for the building is 4 hours, what will be the temperature inside the building at 2:00 PM? At 6:00 PM? When will the temperature inside the building reach 27⁰C? (Answers: 28.3⁰C, 32.5⁰C, 1:16 PM)

The way I see it, the problem is classified as a heating problem, right? Since the aircon is turned off and it's expected the place will heat up, considering the ambient temperature is 35⁰C. Therefore, the differential equation that I should use is dT/dt = k[T-T(ambient)], and the working equation should be T(t) = 35-11e^(t/4). But my answer for T(2) is 16.86⁰C, rather than the correct answer, which is 28.3⁰C. So, I tried to use the cooling equation dT/dt = -k[T-T(ambient)] and the working equation T(t) = 35-11e^-(t/4) instead, and I got the right answers this time.

So, my question is: do I really have to use dT/dt = k[T-T(ambient)] for heating problems or will dT/dt = -k[T-T(ambient)] be okay for it? I'm very confused on which equation to use, since we're taught that there are two equations for Newton's law of cooling, which are the heating and cooling equations above.

Thanks in advance for the answers!

Tags
Newton's Law of Cooling
Heating

Jhun Vert Sun, 10/15/2023 - 10:04

With $k$ being positive at the differential equation, the answer is indeed $28.3^{\circ}\text{C}$ and the value of $k$ in the calculations is negative. I did not solve the problem with initially $-k$ at the differential equation, but I guess the boundary conditions will out put a positive value $k$ if done this way. Here is my solution and tell me what you think.
 

 

And for anyone who stumble at this page, and got stuck with this statement from the problem:

If the time constant for the building is 4 hours

Here is a .pdf file from Bowling Green State University of Bowling Green, Ohio. It says that

the common method of determining the time constant is to see how long it takes for the difference between the current and surrounding temperature to fall to 37% of its initial value.