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Develop an expression for Temp as a function of time

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    A solid body at initial temperature T0 is immersed in a bath of water at initial temperature Tw0. Heat is transferred from the solid to water at a rate [itex]\dot{Q}[/itex]=K[itex]\bullet[/itex](Tw-T), where K is a constant and Tw and T are instantaneous values of the temperatures of the water and solid. Develop an expression for T as a function of time t. Check your result for the limiting cases, t=0 and t=∞. Ignore effects of expansion or contraction, and assume constant specific heats for both water and solid.

    2. Relevant equations[/b[STRIKE]][/STRIKE]

    [itex]\dot{Q}[/itex]=K[itex]\bullet[/itex](Tw-T)

    d(mU)cv/dt=-[itex]\dot{Q}[/itex]

    Cv=dU/dT

    3. The attempt at a solution

    -K[itex]\bullet[/itex](Tw-T)=m*dU/dt

    K[itex]\bullet[/itex](T-Tw)=m*Cv*dT/dt

    dT/dt=K/(m*Cv)(T-Tw)

    Now I will attempt integrating factor

    dT/dt=c1(T-Tw)

    [itex]\mu[/itex](t)*dT/dt=[itex]\mu[/itex](t)*c1(T-Tw)

    [itex]\mu[/itex](t)=exp(∫-c1*dt)=exp(-c1*t)

    d[T*exp(-c1*t)]/dt=exp(-c1*t)*c1(T-Tw)

    T*exp(-c1*t)=∫exp(-c1*t)*c1(T-Tw)dt

    T*exp(-c1*t)=-exp(-c1*t)(T-Tw)

    T=-(T-Tw)

    I would expect a the Temperature of the solid to decrease, as stated in the problem, but eventually level off at an asymptote as it approaches equilibrium.
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 30, 2012 #2
    Maybe this should have been posted in math or introductory physics. can staff move it?
     
  4. Jan 30, 2012 #3
    As written, the equation is not separable. I attempted to use an integrating factor, but I am not sure if it is in the correct form. I obtained an equation that did not contain temperature at all, which is not what the question asks for.
     
  5. Jan 30, 2012 #4

    vela

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    Let's assume K>0. If heat moves from the solid to the water, you must have T>Tw, so ##\dot{Q} = K(T_w-T) < 0##.

    As the solid cools off, you want dU/dt < 0, so it looks like you have an extra negative sign. Your differential equation should be
    \begin{align*}
    m\frac{dU}{dt} &= K(T_w-T) \\
    mc_v \frac{dT}{dt} &= K(T_w-T)
    \end{align*}You're not ready to try solve this yet though because you don't know how Tw varies with time. You need to come up with another differential equation describing its behavior over time.
     
  6. Jan 30, 2012 #5
    [itex]\frac{dTw}{dt}[/itex]=-[itex]\frac{dT}{dt}[/itex]
     
  7. Jan 30, 2012 #6
    m*Cv*[itex]\frac{dT}{dt}[/itex]=m'*Cv'*[itex]\frac{dTw}{dt}[/itex]
     
  8. Jan 30, 2012 #7

    vela

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    No, that's not correct. Try again.
     
  9. Jan 30, 2012 #8
    Check my last post.
     
  10. Jan 30, 2012 #9
    m*Cv*[itex]\frac{dT}{dt}[/itex]=m'*Cv'*[itex]\frac{dTw}{dt}[/itex]
     
  11. Jan 30, 2012 #10

    vela

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    Oops, missed that. Looks okay except you're missing a minus sign.
     
  12. Jan 30, 2012 #11
    [itex]\frac{dTw}{dt}[/itex]=[itex]\frac{m'Cv'}{K(Tw-T)}[/itex]
     
  13. Jan 30, 2012 #12
    Yo, I'm not so good at this. What do I need to work on? Linear system of differential equations?
     
  14. Feb 1, 2012 #13

    vela

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    This is obviously wrong.

    First, find an equation analogous to ##mc_v \frac{dT}{dt} = K(T_w-T)## for Tw. Once you have that, then you need to solve a system of linear differential equations.

    Alternately, you could fix this equation
    $$ mC_v\frac{dT}{dt} = m'C'_v\frac{dTw}{dt}, $$ integrate it, and use the result to eliminate Tw from your first differential equation.

    In any case, it would help if you could explain what you're thinking. Right now, from your posts, it seems like you're just guessing. When you do that, people here aren't inclined to help as much.
     
    Last edited: Feb 1, 2012
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