# What's the temperature distribution of a plane wall insulated on one side with no internal heat gen.?

Let's say we have a plane wall that is insulated on one side, exposed to a fluid on the other side, and has no internal heat generation. Would the shape of the temperature distribution be a constant? I ask because I dont think it would be but a derivation says otherwise so I would like someone to tell me if I am doing this wrong.

The general temperature distribution is created from the heat diffusion equation $$\frac{d^{2}T}{dx^{2}}=0$$ $$\frac{dT}{dx}=C_{1}$$ $$T(x)=C_{1}x+C_{2}$$

I would assume that at the insulation (x=0), the insulation is a adiabatic surface $$\frac{dT}{dx}\bigg|_{x=0}=0$$ So $C_{1}=0$ and $T(x)=C_{2}$ where C2 is the temperature of the surface exposed to the fluid. This is saying that the temperature throughout the wall is the same as the temperature at the surface, but this intuitively doesnt seem right.

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I personally haven't seen the 'heat diffusion equation', though I think it is derivable from the conductivity equation.

Your intuition is misleading you. Whenever there is a temperature gradient, there is a flow of heat, governed by $$\dot{Q}=-kA\frac{\Delta T}{\Delta x}$$ This equation is a special case of Fourier's law. A flow of heat is not possible here, due to the insulator. There will be a little amount of heat flow initially to bring the system to thermal equilibrium, but no more.

The reason you're confused here is that you expect it to have a gradient, based on everyday experiences (which do not involve perfect insulators). The issue here is that we can't see a heat flow, so it is not immediately obvious that a temperature gradient causes a heat flow (and vice versa). Over here, you have a hot liquid on one side, but it is not necessary that a hot liquid gives out heat continuously. It will give out heat till the wall attains the same temperature, after which there will be no net flow of heat.

Over here I am assuming that the top and bottom of the wall are insulated (based on the 'no sources/sinks of heat'. If not, then there will be a temperature gradient, as heat flow will be enabled.

Another way of lookng at these types of questions is by paradigming electricity. Look at temperature differences as potential differences, and heat flow as current ($k$ becomes the electrical conductivity then). Imagine a battery. Connect one end to an insulator, and the other end to a resistor (this is the wall). The insulator and resistor should be now connected. Obviously, you don't get a potential difference across the resistor, as there is no current. Same concept here.

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So my intuition is wrong, but is my derivation correct in this case? Would the temperature of the wall be a constant temperature that is equal to the temperature of the flowing hot/cold liquid? And yes this problem was assuming that it is not insulated on the top or bottom. There is only insulation on one side and the other side is exposed to the fluid. I guess you could say it is an infinitely long/high wall – Greg Harrington Feb 20 '12 at 6:19
@GregHarrington the final result of the derivation npseems correct, but I've never heard of this heat diffusion equation (if you could provide a link, that would help). To me, it seems that $\frac{d^2T}{dT^2}=0$ only for constant heat flow, which is possible at steady state. Since we're dealing with steady state, it seems fine. The adiabatic surface equation will only hold if there is not alternative path for heat(there isn't one here so its fine). So I guess the derivation is correct (It seems to me to be just a longer way of applying Fourier's law) – Manishearth Feb 20 '12 at 6:40
Yep. The wall temperature would be the same as the flowing liquid, by the zeroeth law of thermodynamics. Note that all of this is at steady state. There will be a T gradient till the system reaches steady state. Ok, an infinite wall also works, as long as your water body is also infinite. – Manishearth Feb 20 '12 at 6:42