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Math Help - Temperature Problem

  1. #1
    No one in Particular VonNemo19's Avatar
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    Temperature Problem

    Hi. I'm doing my homework problem set and I came across this one:

    A small metal bar, whose initial temperature is 20^\circ{C}, is dropped into a large container of boiling water. How long will it take the bar to reach 90^\circ{C} if it is known that its temperature increases 2^\circ{C} in 1 second? How long will it take the bar to reach 98^\circ{C}?

    So, I'm sure that we use Newton's cooling law which states that the rate at which the temperature of an object increases or decreases is proportional to the difference between the object and the surrounding temperature, or

    \frac{dT}{dt}=k(T-T_m)\text{ where }k<0.

    So, from the problem, I can infer that (1) T(0)=20; (2) T(1)=22; and (3) T_m=100 (because this is the boiling point of water).

    Newton's differential equation is seperable, so...

    \int\frac{dT}{T-100}=k\int{dt} implies that T(t)=ce^{kt}+100

    Now, T(0)=20\Rightarrow{ce}^{k(0)}+100=20\Rightarrow{c}  =-80

    To find k we use the fact that T(1)=22. So, {-80e}^{k(1)}+100=22.

    All I want to know is if I've done anything wrong so far. So?
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  2. #2
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    Quote Originally Posted by VonNemo19 View Post
    Hi. I'm doing my homework problem set and I came across this one:

    A small metal bar, whose initial temperature is 20^\circ{C}, is dropped into a large container of boiling water. How long will it take the bar to reach 90^\circ{C} if it is known that its temperature increases 2^\circ{C} in 1 second? How long will it take the bar to reach 98^\circ{C}?

    So, I'm sure that we use Newton's cooling law which states that the rate at which the temperature of an object increases or decreases is proportional to the difference between the object and the surrounding temperature, or

    \frac{dT}{dt}=k(T-T_m)\text{ where }k<0.

    So, from the problem, I can infer that (1) T(0)=20; (2) T(1)=22; and (3) T_m=100 (because this is the boiling point of water).

    Newton's differential equation is seperable, so...

    \int\frac{dT}{T-100}=k\int{dt} implies that T(t)=ce^{kt}+100

    Now, T(0)=20\Rightarrow{ce}^{k(0)}+100=20\Rightarrow{c}  =-80

    To find k we use the fact that T(1)=22. So, {-80e}^{k(1)}+100=22.

    All I want to know is if I've done anything wrong so far. So?
    Looks good. An easy way to test it is once you have the DE take the limit as t goes to infinity. In this case, the value better be 100.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dwsmith View Post
    Looks good. An easy way to test it is once you have the DE take the limit as t goes to infinity. In this case, the value better be 100.
    Oh, Ok. That's a good way to look at it. So, you mean to say that once I've found a a function y=\phi which is a solution to the DE, to take \displaystyle\lim_{x\to\infty}\phi(x)?
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    Oh, Ok. That's a good way to look at it. So, you mean to say that once I've found a a function y=\phi which is a solution to the DE, to take \displaystyle\lim_{x\to\infty}\phi(x)?
    Since your equation is T(t), I would label it as \displaystyle\lim_{t\to\infty}T(t)
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  5. #5
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    You said that:

    it is known that its temperature increases in second.

    Why then you need DE ?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by zzzoak View Post
    You said that:

    it is known that its temperature increases in second.

    Why then you need DE ?
    I thought that at first, too, but then I realized that just because it was observed that the temp went up 2 degrees in the first second didn't mean that this increase would be constant. The facts that this was a problem in cooling/heating and that I just read the section about Newton's DE gave it away too.
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  7. #7
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    If it is in the first second then I was wrong. Sorry.You need this DE.
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