Deriving the time needed until Thermal Equilibrium

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?

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?

• You beat me to it!
– Gert
Feb 25, 2021 at 16:13
• @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?
– Phy
Feb 25, 2021 at 17:00
• @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.
– Gert
Feb 25, 2021 at 20:45
• @Phy Now give that man an upvote, like I have done: his answer is 100 % correct.
– Gert
Feb 25, 2021 at 20:47
• 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. Feb 25, 2021 at 22:08

The correct value of the integral with respect to Q, taking into account the integration limit at Q = 0, t = 0 is: $$-\frac{kA}{L}\left[\frac{1}{m_1C_1}+\frac{1}{m_2C_2}\right]t=\ln{\left(\frac{C_1m_1(Q+E_2)+C_2m_2(Q-E_1)}{C_1m_1E_2-C_2m_2E_1}\right)}$$

• 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.
– Phy
Feb 27, 2021 at 22:40
• Really? It seems obvious to me. It's of the form dQ/(a+bQ) Feb 27, 2021 at 23:24
• 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!
– Phy
Feb 27, 2021 at 23:40