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Temperature question

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data

    The temperature, T, in ºC, of a yam put into a 200ºC oven is given as a function of time, t, in minutes, by

    T=a(1-e-kt)+b

    a. If the yam starts at 20ºC, find a and b.
    b. If the temperature of the yam is initially increasing at 2ºC per minute, find k.

    2. Relevant equations

    Given in problem.

    3. The attempt at a solution

    I'm not really sure how to set up this problem to do part a. I think if I had a and b I'd be able to do part b with some derivatives and algebra but I just can't figure out how to get there. My instincts are telling me to set T = 20 = a(1-e-kt)+b but I don't know how to approach this from here.
     
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  3. Nov 4, 2013 #2

    Qube

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    For part a set time = 0. What happens to e^-kt then? Can you solve for one of the constants?
     
  4. Nov 4, 2013 #3
    When t=0, e=1 and therefore a=0. Therefore b=initial T which in the case of a is 20. Ok great. But what do I do to get a? I think I need to set T=20 but I don't know what I need to set t equal to. I know I can't use zero, but aside from that I don't know what to do.
     
  5. Nov 4, 2013 #4

    Qube

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    Generalizing your statement, you're saying that if any number x is multiplied by zero, that number x is zero.

    Do you see a problem with this?

    If you have x*0, does this imply that x = 0?

    Also, is e a constant or a variable?
     
  6. Nov 4, 2013 #5
    I'm saying that at t=0, a(1-e-kt) will equal zero because e0=1, 1-1=0 and a*0=0.

    I wasn't trying to imply that.

    e is the constant Euler's number.
     
  7. Nov 4, 2013 #6

    Qube

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    a*0 indeed equals 0, but we're deviating from the original problem. Are there other parts to the original equation even when you substitute 0 in for time and 20 degrees C in for the temperature?

    Right, and e = 2.71 (approximately). Not 1.
     
  8. Nov 4, 2013 #7
    What do you mean by "other parts"?

    But if we're setting t=0, it will be e-k*0. -k*0=0 so it will be e0=1.
     
  9. Nov 4, 2013 #8

    Qube

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    What happened to the variables you were solving for? Surely if they all canceled out then this would be a bad question on the part of the teacher, right?

    e^0 = 1.

    You stated e = 1.
     
  10. Nov 4, 2013 #9
    Wait what? What do you mean "what happened to the variables you were solving for?" I was only trying to say everything but b is canceled out when t=0. Surely you must see that.

    Ok that was a false. Take it as e0=1
     
  11. Nov 4, 2013 #10

    Dick

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    Ok. So you've got b. Now what happens if t is very, very large?
     
  12. Nov 4, 2013 #11

    Qube

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    T = a(y) + b, where y is substituted in for whatever function that a was multiplied by.

    If y = 0, then

    T = a(0) + b

    Do you see what happens (contrary to what you think?)
     
  13. Nov 4, 2013 #12
    When t gets very large, e-kt approaches 0. So the equation becomes a(1-0)+b=T. This is simplified to a+b=T where b is the initial temperature and T is the final temperature. Therefore a=T-b.

    Yes I get it. But in this scenario a(0)=0. I just simplified it.
     
  14. Nov 4, 2013 #13

    Dick

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    Yes, that's it. t=0 gives you one constant and t=infinity gives you the other one. Now you just need to find k. You are given dT/dt=2 at t=0. So?
     
  15. Nov 5, 2013 #14
    Solving for k I get 1/90, which appears to be the right answer in the back of the book. Along with my answers for a and b. Thanks.
     
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