# Newton's Law of Cooling [closed] 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 ^^

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

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## 1 Answer

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.