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Two materials of mass $m_1$ and $m_1$, specific heat capacity $C_1$ and $C_2$ and temperature $T_1 > T_2$ are separated by a conductive solid with thickness $L$, area $A$ and heat conductivity $k$ which stays fixed regardless of the changing temperature gradient $T_1-T_2=\Delta T$ over time. The amount of energy $Q$ transferred per second is then equal to: $$\frac{dQ}{dt}=\Delta T\cdot\frac{k\cdot A}{L}$$ I am trying to derive the time needed until thermal equilibrium $T_{eq}$ has been reached. Since the energy transfer rate $Q$ depends on $\Delta T$ which in turn changes over time, I would therefore reason that one should calculate the time needed to transfer an infinitesimally small energy $dQ$ and sum that up until the total energy $Q_{eq}$ has been transferred. This would lead to an integration which gives the total time needed: $$t_{eq}=\int_0^{Q_{eq}}\frac{1}{\Delta T\cdot \frac{k\cdot A}{L}}\cdot dQ=\int_0^{Q_{eq}}\frac{1}{\bigg(\frac{E_1-Q}{C_1m_1}-\frac{E_2+Q}{C_2m_2}\bigg)\cdot \frac{k\cdot A}{L}}\cdot dQ$$ Where $E_1$ and $E_2$ are the initial thermal energies of the materials. The term within brackets in the denominator is $\Delta T$ in terms of energy in steps of $dQ$ during the integration. Rewriting the integration in terms of $d\Delta T$ which is equal to $d\Delta T = -dQ\bigg(\frac{1}{C_1m_1}+\frac{1}{C_2m_2}\bigg)$: $$t_{eq}=\int_0^{T_1-T_2}\frac{1}{\Delta T\bigg(\frac{1}{C_1m_1}+\frac{1}{C_2m_2}\bigg)\cdot \frac{k\cdot A}{L}}\cdot d\Delta T$$ I have 2 questions about this approach:

1. The integrations are divergent because the denominator approaches 0, which makes it impossible to derive a definite integral for the time $t_{eq}$. How should one solve this problem?

2. Knowing that the upper limits $Q_{eq}= \frac{E_1C_2m_2-E_2C_1m_1}{C_1m_1+C_2m_2}$ and $T_1-T_2=E_1C_1m_1 - E_2C_2m_2$, the integrations in terms of $Q$ and $\Delta T$ should give the same outcome. However, when calculating the integrations numerically, they give completely different values. How come?

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2 Answers 2

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The equations that should be used are $$m_1C_1\frac{dT_1}{dt}=\frac{kA}{L}(T_2-T_1)$$ and $$m_2C_2\frac{dT_2}{dt}=\frac{kA}{L}(T_1-T_2)$$So, $$\frac{d(T_1-T_2)}{dt}=-\frac{kA}{L}\left[\frac{1}{m_1C_1}+\frac{1}{m_2C_2}\right](T_1-T_2)$$The solution to this equation is $$(T_1-T_2)=(T_1-T_2)_{t=0}\exp{\left(-\frac{kA}{L}\left[\frac{1}{m_1C_1}+\frac{1}{m_2C_2}\right]t\right)}$$What does this tell you about how long it would take for this system to reach equilibrium? On a practical basis, can you estimate how long it would "effectively" take for this system to reach equilibrium?

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  • $\begingroup$ You beat me to it! $\endgroup$
    – Gert
    Feb 25, 2021 at 16:13
  • $\begingroup$ @Chet Miller Thanks. Your 2nd equation is indeed what I deduced but I don't know how you came to the solution, could you please elaborate? Also, shouldn't I then solve $$0=(T_1-T_2)_{t=0}\exp{\left(-\frac{kA}{L}\left[\frac{1}{m_1C_1}+\frac{1}{m_2C_2}\right]t\right)}$$ to calculate the time needed to reach equilibrium? But then $t$ would have to be infinite, why isn't it possible to calculate it this way? $\endgroup$
    – Phy
    Feb 25, 2021 at 17:00
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    $\begingroup$ @Phy But then t would have to be infinite, why isn't it possible to calculate it this way? Because that time really is INFINITE! That's how exponential decay works: $\Delta T$ evolves asymptotically to $0$ for $t\to +\infty$. But you can calculate a time to reach $0.05 \times \Delta(t=0)$ for example. $\endgroup$
    – Gert
    Feb 25, 2021 at 20:45
  • $\begingroup$ @Phy Now give that man an upvote, like I have done: his answer is 100 % correct. $\endgroup$
    – Gert
    Feb 25, 2021 at 20:47
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    $\begingroup$ Mathematically, it is infinite. But, on a practice basis, you can say that equilibrium is attained when you have reached 1% or 0.1%, or 0.01% (or whatever pleases you best) of the initial temperature difference. The differential equation is of the form $\frac{dy}{dt}=-\lambda y$ or $\frac{d\ln{y}}{dt}=-\lambda$. This is a relatively straightforward equation to solve. $\endgroup$ Feb 25, 2021 at 22:08
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The correct value of the integral with respect to Q, taking into account the integration limit at Q = 0, t = 0 is: $$-\frac{kA}{L}\left[\frac{1}{m_1C_1}+\frac{1}{m_2C_2}\right]t=\ln{\left(\frac{C_1m_1(Q+E_2)+C_2m_2(Q-E_1)}{C_1m_1E_2-C_2m_2E_1}\right)}$$

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  • $\begingroup$ Yes! This is indeed the solution for my integral in terms of $Q$. Thank you so much for your time and effort. I was unawarely considering the antiderivatives as the solutions the whole time, forgetting that the integrals haves specified boundaries. I am not sure how you have managed to solve this integral in terms of $Q$ though, many calculation tools on websites don't seem to be able to solve it. $\endgroup$
    – Phy
    Feb 27, 2021 at 22:40
  • $\begingroup$ Really? It seems obvious to me. It's of the form dQ/(a+bQ) $\endgroup$ Feb 27, 2021 at 23:24
  • $\begingroup$ Never mind, I just noticed that rewriting $\ln(\frac{\Delta T}{\Delta T_{t=0}})$ from your very first solution in terms of $Q$ also gives the same outcome. Thanks again! $\endgroup$
    – Phy
    Feb 27, 2021 at 23:40

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