Results 1 to 7 of 7

Math Help - need help with differential (problem based)

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    17

    need help with differential (problem based)

    The formula [tex] \frac{dT}{dt}=k(T-Tm)[/math] where k is a constant of proportionality also holds when an object absorbs heat from the surrounding medium. If a small metal bar whose initial temperature is 20oC is dropped into a container of boiling water, how long will it take the bar to reach 90oC if it is known that its temperature increased 20oC in 1 second? ? How long will it take the bar to reach 98oC ?


    i'm really clueless on how to solve this problem. can anyone help? tq

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    This is Newton's Law of Cooling.

    We have the initial conditions T(0)=20, \;\ T(1)=40

    Water boils at 100 degrees C, so T_{m}=100. That is the temperature of the surrounding medium. In this case, the surrounding medium is boiling water.

    \frac{dT}{dt}=k(T-100)

    Separate variables:

    \frac{dT}{T-100}=kdt

    Integrate:

    ln(T-100)=kt+c

    T-100=e^{kt+c}

    T-100=Ce^{kt}

    Now, use the initial conditions to find k and C.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2008
    Posts
    17
    Why do you use Tm=100 instead of T=100?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Because T_{m} is the temp. of the boiling water and T is the temperature at any time t.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2008
    Posts
    17
    i've tried solving but ended up getting ridiculous value for the answer. can you show me the steps? tq
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The worst part is over. You have two equations with two unknowns.

    This is where you use the initial conditions you were given.

    We have:

    40=100+Ce^{k(1)}

    20=100+Ce^{k(0)}

    Solve for k and C.

    From the second equation we can clearly see that C=-80

    Then we get, upon subbing into the first:

    40=100-80e^{k}

    k=ln(3/4)

    So, our equation is T=100-80e^{ln(3/4)t}

    or T=100-80(3/4)^{t}

    To find how long it takes to reach 98C, set T=98 and solve for t.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2008
    Posts
    17
    -
    Last edited by bleu90; September 6th 2008 at 06:15 AM. Reason: false claim
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Indices based problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 9th 2010, 11:51 AM
  2. Help: Poisson-based problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 19th 2009, 02:39 PM
  3. Problem based on ratio
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 31st 2009, 09:25 AM
  4. Problem based on square root
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 30th 2009, 09:39 PM
  5. Problem based learning
    Posted in the Geometry Forum
    Replies: 6
    Last Post: January 18th 2006, 02:34 PM

Search Tags


/mathhelpforum @mathhelpforum