## Something About My Deduction On the Fourier's Law of Heat Conduction

My hypothesis is to use two different stable heat sources with different tempreature T1 and T2 (T1>T2) transmits the heat . Then I let the distance between this two heat sourse filled with idea gas or ideal metal in a tube. So if the distance is L, the heat capacity is Cv (Constant). So the temperature at r. (r is the distance to heat source T2 0<r<L). Then when the system balances, we have the temperature T(r)=T1+(T2-T1)r/L. That is the fourier's law form. I will tell you my way to deduce this formula in details if you are interested in this research.

Do you guys think I can analyze the heat in this way: Seperating the heat in the tube into many chunks of heat units ΔV. The volume of each units is the same. Then as the area S is constant. So the length of each unit will be the same Δl.

Here is the hypothesis; if
I. The heat units are all continuous
II. The volume of each units cannot be compressed

Then I will have each unit has the same velocity v0

And other hypothesis, if
in a very short time t, a heat ΔQ(t) transmits into the tube, each part in the tube will obsorb the same amount of heat (ΔL/L)ΔQ(t).

Do you agree with all these hypothesis? If so, I can deduce the Fourier's Law of Heat Conduction. If not, please tell me your thoughts. Thank you!
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Recognitions:
Homework Help
 Quote by Zhaozhong Shi My hypothesis is to use two different stable heat sources with different tempreature T1 and T2 (T1>T2) transmits the heat.
Heat will flow from the hot place to the cold one. This makes the cold one hotter - in order for the cooler part to stay at T2, it would have to gradually crank down the rate it pumps heat until it removes it as fast as heat arrives.
 Then I let the distance between this two heat sourse filled with idea gas or ideal metal in a tube.
Gas or metal - it makes a difference. For instance, the gas will flow by convection between the hot and cold parts of the tube.
 So if the distance is L, the heat capacity is Cv (Constant). So the temperature at r. (r is the distance to heat source T2 0
At equilibrium - I guess...
 That is the fourier's law form. I will tell you my way to deduce this formula in details if you are interested in this research. Do you guys think I can analyze the heat in this way: Seperating the heat in the tube into many chunks of heat units ΔV. The volume of each units is the same. Then as the area S is constant. So the length of each unit will be the same Δl.
If you are careful sure.
 Here is the hypothesis; if I. The heat units are all continuous II. The volume of each units cannot be compressed Then I will have each unit has the same velocity v0
What do you mean by velocity? You mean the heat flow through each length interval dl will be the same rate?

 And other hypothesis, if in a very short time t, a heat ΔQ(t) transmits into the tube, each part in the tube will obsorb the same amount of heat (ΔL/L)ΔQ(t).
You mean the tube warms up - but the hot end of the tube will be a higher temperature than the cold part (conditions due to your setup). Is this consistent with all parts having the same stored heat?

But what you are talking about sounds like the normal descriptions of heat conduction. If you can get it to do work on the way, then you have a heat engine.
 [QUOTE=Simon Bridge;3954811]Heat will flow from the hot place to the cold one. This makes the cold one hotter - in order for the cooler part to stay at T2, it would have to gradually crank down the rate it pumps heat until it removes it as fast as heat arrives. [QUOTE/] I suppose the heat source have A LOT OF mass so that the heat capacity is so big that the temperature will not change. [QUOTE=Simon Bridge;3954811]Gas or metal - it makes a difference. For instance, the gas will flow by convection between the hot and cold parts of the tube. Yes, I am wrong. I ignored when the temperature of the gas increases, its pressure will increase at the same time. I will use the ideal metal instead. [QUOTE=Simon Bridge;3954811]At equilibrium - I guess... Yes, it is. But let's imagine the heat inputs differently from heat source T1(t) to T2(t). That means it is a function of time ΔQ(t). It will make a difference. [QUOTE=Simon Bridge;3954811]What do you mean by velocity? You mean the heat flow through each length interval dl will be the same rate? Yes, I think each heat unit flows through each length interval ΔL will be the same rate. [QUOTE=Simon Bridge;3954811]You mean the tube warms up - but the hot end of the tube will be a higher temperature than the cold part (conditions due to your setup). Is this consistent with all parts having the same stored heat? First, let's say a heat units Q1 enters on end of the tube. Then when it leaves from the other end of the tube, it becomes Q2. So the heat the tube obsorb will be Q1-Q2. Let's hypothesize that each part obsorb the same heat (ΔL/L)ΔQ at DIFFERENT TIME. Here is my description: Let's imagine. When t=0, the no heat flows in the tube. So every part in the tube is the same temperature T2 When t=Δt, a heat unit ΔQ(Δt) flows in the tube; it is at the point A (ΔL) (Assume that ΔL=v0Δt, v0 is a constant) heat. Let's say it obsorbs Δq(Δt), so the temperature at A will be T2+Δq/Cv (Cv is heat capacity). Then due to the temperature difference, the heat ΔQ(t)-Δq(t) will transfer to point B (2ΔL) and the other part of the tube still remains T2. When t=2Δt, another heat ΔQ(2Δt) arrives at point A. If point A obsorbs Δq(2Δt), totally obsorbing Δq(Δt)+Δq(2Δt), then the temperature at point A will be T2+(Δq(Δt)+Δq(2Δt))/Cv the heat ΔQ(Δt)-Δq(t) arrives at point B. At the same time, if ponit B oborbs the same heat as point A, then temperature of point B will be T2+Δq(Δt)/Cv. Then due to the temperature difference, the heat ΔQ(t)-2Δq(t) will transfer to point C (3ΔL) and ΔQ(2Δt)-Δq(2Δt) will transfer to point B. When t=nΔt, the original heat ΔQ(Δt) passed through the end of the tube. At point A (ΔL), we will have the total heat it has obsorbed, q(A)=Δq(Δt)+Δq(2Δt)+Δq(3Δt)+...+Δq(nΔt)=∫(dq/dt)dt The region (0,nΔt) So point B (2ΔL) has obsorbed, q(B)=Δq(Δt)+Δq(2Δt)+Δq(3Δt)+...+Δq((n-1)Δt)=∫(dq/dt)dt The region (0,(n-1)Δt) Then at point P (kΔL), it has obsorbed, q(P)=Δq(Δt)+Δq(2Δt)+Δq(3Δt)+...+Δq((n-k-1)Δt)=∫(dq/dt)dt The region (0,(n-k-1)Δt) [QUOTE=Simon Bridge;3954811]But what you are talking about sounds like the normal descriptions of heat conduction. If you can get it to do work on the way, then you have a heat engine. If you guys agree with my hypothesis, I will tell you how to deal with the sums above. Thanks

## Something About My Deduction On the Fourier's Law of Heat Conduction

Any ideas?
 @Simon Bridge I need your help

Recognitions:
Homework Help
 I suppose the heat source have A LOT OF mass so that the heat capacity is so big that the temperature will not change.
So this is the classical picture of a heat engine ... you have a source and a sink, a hot and cold heat reservoir, always at the same temperature and something to conduct the heat between them.

 When t=0, the no heat flows in the tube. So every part in the tube is the same temperature T2
The entire tube starts at thermal equilibrium with the cold reservoir.

Note - if the tube absorbs heat as described then it will eventually be the same temperature as the hot reservoir. Usually the metal rod will be hottest by the hot reservoir and coldest at the cold reservoir ... and have a constant temperature gradient down it's length.

That would be at steady state - so heat is leaving the tube as fast as it is added.

In the transient state, you get the normal rules for heat conduction.

 Recognitions: Homework Help You have to decide if your reservoirs are constant temperature OR provide a constant heat flow. This is much how you'll be used to handling electric circuits where the power source can either be a voltage or a current source but not both. Just like we usually have constant-voltage supplies, we usually have constant temperature sources (in idealized systems of course). If you bring a hot reservoir in contact with a rod, you can compute the heat flow depending on the temperature of the rod and it's thermal conducting properties. If ΔQ is the same for every Δt then you can say the heat current, ΔQ/Δt, is a constant. But this situation will only apply a long time after the system is first hooked up. Since you have a simple geometry it is easy to show that$\frac{\Delta Q}{\Delta t} = -kA\frac{\Delta T}{\Delta L}$ which makes your last relation more like: $$\Delta T = -\frac{1}{kA}\frac{\Delta Q}{\Delta t}\Delta L$$ ... if the current is a constant then this collapses to what you got: but all it tells you is that the temperature gradient is a constant. Your conclusion seems to agree with mine but your method appears a tad confused. So what is it that you are trying to work out?