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Cooling of a cup of hot chocolate

  1. Sep 10, 2017 #1
    1. The problem statement, all variables and given/known data
    A cup of hot chocolate of radius r and height H is completely full. The density ρ and specific heat C of chocolate are known. The cup is very well insulated so that heat is only exchanged with the outside room from the open top. The chocolate is initially at temperature T0, the outside room is kept at temperature Tr, the temperature of the chocolate at the top and bottom of the cup are called Tt and Tb and they vary with time. Assume the heat flow from the air to the cup is [tex]F_t = - K_t \pi r^2 (T_t - T_r)[/tex]. Assume the heat flow inside the cup is [tex]\vec F = - k_i \pi r^2 \vec \nabla T[/tex]. Note that Kt, ki have different units.

    How do Tt and Tb depend on time?

    2. Where is this problem from?
    I thought of this problem while wondering how long the chocolate on my desk would take to cool. It looked easy at first and when I tried to do it I found it harder than expected. I expected it would just be a matter of writing an energy balance, but I soon found myself deriving the heat equation with what looked like fairly nasty boundary conditions (BCs). I haven't been doing physics in a while so your guidance in solving it would help. Also, do tell me if the assumptions in the problem are nonsensical.

    3. Relevant equations, attempt at a solution
    The following may be irrelevant because if you don't know the heat equation, you won't be able to help me with this, and if you know the heat equation, you probably don't need to see it derived again.

    Take a small layer of chocolate between heights h and h + dh. The energy balance will be (energy in) - (energy out) + (delta internal energy) = 0.

    (energy in) is [tex]- k_i \pi r^2 \frac{\partial T(h, t)}{\partial h}[/tex].
    (energy out) is [tex]- k_i \pi r^2 \frac{\partial T(h + dh, t)}{\partial h}[/tex] in the interior, [tex]- K_t \pi r^2 (T_t - T_r)[/tex] at the top.
    (delta internal energy) is [tex]\rho C \pi r^2 dh \frac{\partial T}{\partial t}[/tex].

    Put the three together and you'll get the heat equation in the interior, a weirder equation at the top. We could just look at this as the heat equation with a messy BC the top. The easiest way to frame this BC for me is to say that the total internal energy changes just because of the exchange at the top: [tex]\frac{\partial}{\partial t} \int_0^H{T(h, t) C \rho \pi r^2 dh} = - K_t \pi r^2 (T(H, t) - T_r)[/tex]. The other BC is [tex]T(h, 0) = T_0[/tex] for any h.
  2. jcsd
  3. Sep 10, 2017 #2
    This is not set up quite correctly. The heat balance on the section between ##h## and ##h+\Delta h## should be:

    [tex]- k_i \pi r^2\left( \frac{\partial T}{\partial h}\right)_h-\left[- k_i \pi r^2 \left(\frac{\partial T}{\partial h}\right)_{h+\Delta h}\right]=\rho C \pi r^2 \Delta h \frac{\partial T}{\partial t}[/tex]
    If we divide by ##\Delta h## and take the limit as ##\Delta h## approaches zero, we obtain the transient heat conduction equation:
    $$\rho C\frac{\partial T}{\partial t}=k_i\frac{\partial^2T}{\partial h^2}$$

    The boundary condition at the top, ##h=H## is: [tex]- k_i \pi r^2 \left(\frac{\partial T}{\partial h}\right)_{h=H}=K_t \pi r^2 (T - T_r)[/tex]or[tex]k_i \left(\frac{\partial T}{\partial h}\right)_{h=H}=-K_t (T - T_r)[/tex]
  4. Sep 10, 2017 #3
    Thank you. I thought about it for a while and I agree.

    I haven't been solving PDEs for a long time. Could the 1 dimensional heat equation with this type of BC be solved by standard techniques?
  5. Sep 10, 2017 #4
    No problem.
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