# Thermal conductivity

Discussion in 'MCAT Discussions' started by nimish, Apr 12, 2007.

1. ### nimish

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Can someone please explain how it is possible for the rate of heat flow to be equal in slabs of different lengths?
The formula for heat flow is Q/t = kA[(Th-Tc)/L]
This formula clearly shows that heat flow is inversly proportional to the length of the material, yet on page 47 of EK Chemistry, they say "... the rate of heat flow, Q/t, would be the same in all slabs even if they each had different lengths..."

How is that possible?
A brief explanation would be greatly appreciated!

3. ### scottj72

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I think the key word is "rate". The rate would not change no matter how long the material because it would be like any other physical property. A good example similar is specific heat. What I think you were thinking about is the total time. This would change with the length of the material. However, both of these values are in the denomenator, so believe that would make those two values directly proportional. So any change in length would change the time directly thus the "rate" remains unchanged.

Hope that helps!

4. ### RSAgator Junior Member

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That's fourier's law of heat transport i believe. I think the problem itself is slightly misleading though. The way the equation is actually written is:

dQ/dt = kA dT/dX

So temperature is a function of distance. That in mind, if the length of the wall changes, then the temperature difference is going to change as well.

Think of it this way:

You have a constant heat source applied on one end of a wall. Heat is going to flow through the wall, and the wall is going to be a certain temperature (Th) on the side next to the heat source, and a different temperature (Tc) on the other end of the wall (the side away from the heat source). If you increase the length of the wall, the temperature on the cold side is going to be colder, so as the length increases, the temperature difference is going to increase as well, and the flow rate into the wall is going to remain the same.

Looking at it a different way, if you have a heat source applied to one side of a wall that sends 10joules of heat into the wall, that heat source isn't going to change if the wall gets longer, it will still send 10 joules of heat into the wall. I hope that helps =)

edit: that's 10 joules/second btw

5. ### nimish

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I understand #2 completely, and it makes sense.
For #1, are you saying that its as simple as this: if you increase the denominator, the numerator will change as well (it will increase/decrease linearly with the denominator; if L goes up, then the difference between the hot and cold also goes up). So the overall net result is that Q/t will remain constant?

If my understanding of this is correct, then thank you guys for your help.
If not.....oh boy

6. ### RSAgator Junior Member

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Exactly. The reason that you're having some trouble understanding it is simply because it is poorly defined. The temperatures are taken at the ends of the wall, where Th corresponds to the temperature at the side where the heat source is, and Tc corresponds to the temperature on the opposite side. It is cooler because as the heat gets "conducted" through the wall, it gets used up to raise the temperature of the wall, so as you get to the end of the wall there is less heat available to heat the wall up, resulting in the cooler temperature.

Thus, as you increase the length of the wall, you decrease the amount of heat available to heat up the far end of the wall, and thus you decrease the temperature of that end. Keep in mind that this is an incredibly simplified view of a potentially complex problem, but the best way to look at it is just with common sense (#2). You basically have a heat source in contact with a heat sink, and nothing you do to the heat sink is going to change the heat generated by the heat source.

If you changed the material of the wall, which would change k in that equation, you would not change the heat generated by the heat source, but you would change the temperatures.

I know i went a bit more in depth than necessary, but to put things in perspective chemical engineers essentially take entire classes based upon this one equation =)

7. ### nimish

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Thank you again for your help and for the explanation, it looks like Thermodynamics won't be a problem for you on the exam!

8. ### superduper12

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Q/t = kA[(Th-Tc)/L]

Situation 1 - comparing lengths slab 1> slab 2

so pretend you have two slabs between the heat source and heat sink. slab 1 and slab 2....slab 1 is longer than slab 2....does this mean that the temperature gradient across slab 1 will be greater than the tempeature gradient across slab 2. I understand that they conduct heat at the same rate

Situation 2 - comparing thermal conductivity slab 1 > slab 2

Now the slabs are the same length but slab 1 has a higher thermal conductivity (k)than slab 2. so heat will flow faster in slab 1. because it flows faster, t increases and...and Q/t remains the same for both slabs

Situation 3 - comparing area slab 1> slab 2

slab 1 has a larger crosssectional area than slab 2. since slab 1 has a greater area, it takes a less time...so in the end Q/t...stays the same for both slabs.

am I thinking of this correctly?...or is time not a factor and I should think about temp gradients.