0
$\begingroup$

Newton's Law of Cooling/Warming

As shown in Figure 3.3.11, a small metal bar is placed inside container A, and container A then is placed within a much larger container B. As the metal bar cools, the ambient temperature $T_A(t)$ of the medium within container A changes according to Newton's law of coding. As container A cools, the temperature of the medium inside container B does not change significantly and can be considered to be a constant $T_B$. Construct a mathematical model for the temperature $T(t)$ and $T_A(t)$, where $T(t)$ is the temperature of the metal bar inside container A. Find a solution of the system subject to the initial conditions $T(0) = T_0 , T_A(0) = T_1$.

what I had try:

$\frac{dTA}{dt} = -k(T_A-T_B)$

$T_A - T_B = Ce^{-kt}$

$T_A(0) - T_B = Ce^{-k0}$

$T_1-T_B = C$

so, $T_A(t) = (T_1-T_B)e^{-kt}$

$\frac{dT}{dt} = -K(T-T_A)$

$\frac{dT}{dt} + KT = KT_A$

$e^{Kt}\frac{dT}{dt} + KTe^{Kt} = KT_Ae^{Kt}$

$\frac{d(e^{Kt}T)}{dt} = KT_Ae^{Kt}$

$Te^{Kt} = K(T_1-T_B)\int{e^{(K-k)t}} + C$

$T = K(T_1-T_B)\frac{e^{(K-k)t}}{(K-k)e^{Kt}} + C$

$T(0) = T_0$

$C = T_0 - \frac{K(T_1-T_B)}{(K-k)}$

so,

$T(t) = K(T_1-T_B)\frac{e^{(K-k)t}}{(K-k)e^{Kt}} + T_0 - \frac{K(T_1-T_B)}{(K-k)} $

but i doubt of my assumption that $\frac{dTA}{dt} = -k(T_A-T_B)$ , i also think that $T_A$ is impacted by $T$ too. help please ^^

$\endgroup$

closed as off-topic by John Rennie, user36790, ACuriousMind, Qmechanic Apr 23 '16 at 21:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Community, ACuriousMind, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

The metal bar cools by conduction to container A, and loses heat at a rate proportional to the temperature difference so

$\frac{dT}{dt}=-k_1(T(t)-T_A(t))$

Meanwhile container A has heat incoming from the metal bar, and is losing heat to container B, so

$\frac{dT_A}{dt}= - k_2(T_A(t)-T_B) + k_1(T(t)-T_A(t)) = - k_2(T_A(t)-T_B) -\frac{dT}{dt}$

Then solve those coupled equations.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.