# Thread: Modeling Heat Loss

1. ## Modeling Heat Loss

I know that this is physics rather than pure mathematics, but this seemed like a better board to get my question answered on.

I have an object at an initial heat of h. It loses heat at a rate of k*h^4 per second, where k is a constant. It also gains heat at g per second.

How can I mathematically model this so that I can find heat at any time past the initial start?

Thanks!

2. It sounds like you'd have the following differential equation (DE):

$dh/dt=-kh^{4}+g$.

To find the heat, you would need to solve this DE.

3. Err, whoops. I may have written the problem incorrectly. Let me rephrase:

I have an object with an initial heat of h0. It loses heat at a rate of k*h^4 per second, where k is a constant and h is the current heat, and gains heat at a rate of g per second.

How can I mathematically model this so that I can find heat at any time past the start point?

4. I think my previous reply still stands. You will use the initial condition $h(0)=h_{0}$ in solving the DE. Do you have a background in solving DE's?

5. I do not have a background in solving differential equations; that may be the problem. This is part of a model of heat loss from a hot air balloon that I am doing for a freshman engineering course, and no one in my group knows differential equations.

6. Originally Posted by Ballooner
I do not have a background in solving differential equations; that may be the problem. This is part of a model of heat loss from a hot air balloon that I am doing for a freshman engineering course, and no one in my group knows differential equations.
This can be a DE but more then likely we are supposed to use decay principles.

$H(t) = h_0 - kh^4 t + gt$

Where t is the time in seconds.

7. Ok. This is a crash course on ODE's: first line of attack: separation of variables. Can you get all the $h$'s on one side, and all the $t$'s on the other? In this case, you can. You obtain, by using multiplication of differentials:

$\frac{dh}{-kh^{4}+g}=dt$.

What you do then is integrate both sides. I would use the definite integral, so that you can more easily take into account the initial conditions:

$\int_{h_{0}}^{h}\frac{dx}{-kx^{4}+g}=\int_{0}^{t}d\tau$,

where I've used the dummy variables $x$ and $\tau$ to avoid confusion. At least, I hope I've avoided confusion.

In performing the integration on the LHS, you're going to get a rather messy expression. You may not be able to solve it explicitly for $h$. That may or may not be a severe drawback, depending on what you want to do with the equation.

Reply to AllanCuz: What would your variable $H(t)$ be representing?

8. Originally Posted by Ackbeet
Ok. This is a crash course on ODE's: first line of attack: separation of variables. Can you get all the $h$'s on one side, and all the $t$'s on the other? In this case, you can. You obtain, by using multiplication of differentials:

$\frac{dh}{-kh^{4}+g}=dt$.

What you do then is integrate both sides. I would use the definite integral, so that you can more easily take into account the initial conditions:

$\int_{h_{0}}^{h}\frac{dx}{-kx^{4}+g}=\int_{0}^{t}d\tau$,

where I've used the dummy variables $x$ and $\tau$ to avoid confusion. At least, I hope I've avoided confusion.

In performing the integration on the LHS, you're going to get a rather messy expression. You may not be able to solve it explicitly for $h$. That may or may not be a severe drawback, depending on what you want to do with the equation.

Reply to AllanCuz: What would your variable $H(t)$ be representing?
In first year engineering we don't go through ODEs so I don't think this will be expected. Though, if you get the right answer why not learn it?

H(t) represents heat at time t. It should be noted that this appears on both sides of the equation, but the question asks us to model the behavior, not isolate H(t) so it is an acceptable as an answer.

9. Hmm. DE's are often introduced in Calc II these days, at least the beginnings of them. I'm not sure I buy your answer, for two reasons.

1. The original problem says, "I have an object with an initial heat of h0. It loses heat at a rate of k*h^4 per second,..." When I see the word "rate", I think derivative.

2. Dimensional Analysis. Again, the quote reads, "I have an object with an initial heat of h0. It loses heat at a rate of k*h^4 per second,..." The units don't work out to have rate on the right-hand-side of the equation (with units of heat per second), and just heat on the left-hand-side.

10. Originally Posted by Ackbeet
Hmm. DE's are often introduced in Calc II these days, at least the beginnings of them. I'm not sure I buy your answer, for two reasons.

1. The original problem says, "I have an object with an initial heat of h0. It loses heat at a rate of k*h^4 per second,..." When I see the word "rate", I think derivative.

2. Dimensional Analysis. Again, the quote reads, "I have an object with an initial heat of h0. It loses heat at a rate of k*h^4 per second,..." The units don't work out to have rate on the right-hand-side of the equation (with units of heat per second), and just heat on the left-hand-side.
ODEs are not part of typical engineering course work for first years, I would know, being an engineering student.

Second,

We have $H(t)$ that has initial heat $h_0$ sustaining a loss of $k H(t)^4 t$ and a gain of $gt$

The equation is,

$H(t) = h_0 - kH(t)^4 t + gt$

Which in words is, the heat at time t is equal to the original heat minus the loss given by $kH(t)^4$ per second plus the heat gain given by $gt$ ...which is exactly what the question states.

We multiply it by time! It's per second multiplied by a given amount of seconds. If you have $per second$ times $x seconds$ what do we get? We get x! Which is a constant....

There is nothing wrong with that

11. We'll just have to agree to disagree, I guess. I'm not following where your time multiplication came from. The problem doesn't mention that. I would agree that heat per second times time will give you units of heat. But there's nothing that I can see in the problem statement to justify that. Is there something else in the context of the problem of which I'm unaware?