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In this problem, there is a water heater, and I am trying to derive the equation for $T$, the temperature of the water, in terms of minutes, $t$. I am given the following information:

  • Heat is supplied at a constant rate of $1200 \ Btu$ (British thermal units) per minute.
  • Heat is lost to the surroundings at a rate, $L$, proportional to the difference between the heater temperature and the room temperature. That is, $L = h(T - 70)$, where $L$ is the loss rate in $Btu/min$ and $T$ is the water temperature. The room temperature is $70$ degrees Fahrenheit, and $h$ is a proportionality constant called the heat transfer coefficient.
  • The heater warms water at a rate, $\frac{dT}{dt}$, proportional to $(1200 - L)$
  • The water would warm up at $3$ degrees Fahrenheit per minutes if there were zero losses to the surroundings.
  • In $10$ minutes the heater warms the water to $96$ degrees from the room temperature of $70$ degrees.

I think this would mean that $\frac{dT}{dt} = (1200 - h(T - 70))$ but I have no idea how to anti-derive that, or if that's even the way to go about it.

Any help would be greatly appreciated. Thanks!

Edit: By separating the differentials, I was able to get $T = \frac{e^{-t-C} - 1200 - 70h}{-h}$, but I'm not sure where to go from here.

Edit 2: Thanks to Simply Beautiful Art I realize I forgot about the "proportional" part of $\frac{dT}{dt}$, which is actually $\frac{1200-L}{400}$. I used this and separated the differentials to solve for $T$, and I got:

$T = \frac{-400}{h}(e^{-th/400+C} -3 + \frac{70h}{400})$

I'm trying to solve for $h$ and $C$ by using a system with the $t = 0$ and $t = 10$ points given in the last bullet point above, but I can't seem to figure out how to solve that.

1 Answers 1

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We know that $\frac{dT}{dt}$ is proportional to $(1200-L)$, so

$$\frac{dT}{dt}=k(1200-L)$$

For some constant $k$. From the next bullet point, we know that if the losses $L$ were zero, then $\frac{dT}{dt}=3$, so

$$3=k(1200-0)\implies k=\frac1{400}$$

Thus, we have

$$\frac{dT}{dt}=\frac{1200-L}{400}$$

And from there you got this?

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    Ah, I forgot about the proportional part, thanks! Now though, I get T = (-400/h)(e^(-th/400+C) -3 + 70h/400); I tried to solve that using a system with the t=0 and t=10 points, but I can't seem to figure out how to solve it with the e and everything. Any ideas? – 2017-02-12
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    @user3316549 Once you isolate the e^() term, just take the natural log of both sides. Solve for $C$ in terms of $h$, and find $h$ through substitution. If worse comes to worse, I think you should use a calculator. – 2017-02-13
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    I got it, thanks so much! – 2017-02-13