# Homework Help: Internal Energy struggle

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1. Nov 26, 2014

### Alettix

Hello,

So I have been struggling to understand internal energy for quite a long time and I hope you will be able to help me get straight with it. The most common definition I find on the internet is "The sum of the kinetic and potential energy of the molecules/atoms in a system" (often gases). Too me, this sounds like a physical quantity bigger than the heat energy, because the latter takes only the kinetic energy of the molecules in account (Am I wrong?). But the equation for internal energy is:
ΔU = ΔQ - W = ΔQ - p × ΔV
where ΔU is the change in internal energy, ΔQ the heat given to the system and W (or pΔV) the work done by the system (often change in volume). So this means that the change of internal energy is smaller than the change in heat. I would be really thankful if somebody could explain the differences between heat and internal energy, and give the proper definitions of them.

What confuses me even more is the following example in my physics book:

1. The problem statement, all variables and given/known data
"A closed glass cylinder has a piston which can be assumed to move very easily and without friction. The cylinder contains 1,0 dm^3 of air with a temperature of 20 degrees Celsius and has a pressure equivalent to 101,3 kPa. The air is then heated to 100 degrees Celsius.
a) How much heat was given to the air?
b) How big is the work done by the system by change in volume?
c) What's the change in internal energy?"

2. Relevant equations and information
ΔU = ΔQ - W = ΔQ - p × ΔV (1)
pV=nRT → V1/T1=V2/T2 (2)
ΔE = mcΔT (3)
specific heat capacity of air c = 1,01 kJ/(kg × K)
ρ = m/V (4)
density of air = 1,293 kg/m^3

3. The solution in the book

a) The heat energy required to increase the air's temperature by 80 degrees is calculated with equation (3), with the help of equation (4) to determine the mass of the air. This gives the answer: 104 J

b) The pressure is assumed to be constant (thanks to the very mobile piston), and so the change in volume can be calculated with equation (2). This value is then multiplied with the magnitude of the pressure (101,3 kPa) and so the work done by the system is determined to be 28 J.

c) With the help of equation (1) the change in internal energy is calculated to be: 104-28= 77 J.

3. What I don't understand and need help with

I understand the calculations, but once again I can't really link it to the definitions for heat and internal energy. What I find the most incomprehensible is how we can say that 104 J go to increase the temperature of the air, when we know that 28 J of these will go to expand the cylinder! I would find it more logical to say that only 77 J go to heat the gas, but then this would lead to the conclusion that heat is equal to internal energy, which is certainly not right according to equation (1).

Big thanks to somebody who can make the definitions of internal energy and heat clear, and explain why we can say that 104 J heat the system when we know that 28 J will go to work!

2. Nov 26, 2014

### Staff: Mentor

It also takes potential energies into account if they come from internal potentials (e.g. between atoms in molecules, not from gravity of earth).
Expanding the cylinder is part of the energy needed "to increase the temperature". There is also the heat capacity for a constant volume which has a different value.

3. Nov 26, 2014

### Alettix

Thank you for your answer mfb! :)

But isn't then the definition for heat and internal energy the same? What's the difference?

I don't understand this, WHY is expanding the cylinder a part of the energy needed "to increase the temperature"?
And do you mean that the book used the wrong value for the heat capacity? Does that affect the problem and its results in any other way than just changing the recieved values for the energies?

Thank you!

4. Nov 26, 2014

### haruspex

No, that is the difference: heat only counts internal (as opposed to bulk) KE, whereas internal energy also counts internal PE.
In the context of the question, the issue is how much heat do you have to supply for the temperature to go up by 80oC. Some of the heat supplied will be spent (as work) expanding the cylinder, but that's all part of what's needed as an external input.

5. Nov 26, 2014

### Staff: Mentor

You might find this FAQ entry helpful: https://www.physicsforums.com/threads/what-is-heat.511174/ [Broken]

Last edited by a moderator: May 7, 2017
6. Nov 26, 2014

### Staff: Mentor

The work and the heat represent exchanges of energy across the boundaries of your system with the surroundings. The internal energy represents the amount of energy that your system contains within its boundaries. Exchanges of heat and work with the surroundings can increase the amount of internal energy or decrease the amount of internal energy within your system.

In your example, heat is added to your system across its boundary, but your system also does work on the surroundings by expanding. So the heat added to your system tends to increase its internal energy, and the work your system does on the surroundings acts to decrease the internal energy of your system. In your problem, the heat added wins out over the work done, and the internal energy increases (but not as much as if all the heat went into increasing the internal energy).

Chet

7. Nov 26, 2014

### Staff: Mentor

What about the internal potential energy in vibrations of molecules? This is part of heat.

8. Nov 26, 2014

### Staff: Mentor

In your problem, 104 Joules of heat Q enter through the boundaries of your system, but not all of this heat is used to increase the temperature of the air. Only 77 of these Joules go into increasing the temperature of the air (corresponding to the change in internal energy ΔU); the remainder of the heat (27 Joules) is used to cause the gas to expand (do work) against the surroundings.

Chet

9. Nov 26, 2014

### haruspex

Ah, ok.

10. Nov 27, 2014

### Alettix

Firstly, thank you for your answer Sir!

Well, this was the way thought about it too! But if only 77 J go to increas the temperature of the air, the increase will only be:
ΔT = ΔE / (m × c) = 77 / (1,293 × 10^-3 × 1010) ≈ 59,0 °K
And so the temperature of the air is: 20+59 = 79 °C, but this is against the statement of the problem, because the temperature should be 100 °C.
How does that work out?

11. Nov 27, 2014

### Staff: Mentor

Realize that the specific heat capacity you are given is the specific heat capacity at constant pressure. As Chestermiller explained, that tells you the amount of heat that must flow to give you a given change in temperature. But it does not tell you the increase in internal energy--for that you must apply the first law of thermo.

12. Nov 27, 2014

### Staff: Mentor

The way this works is as follows: The change in internal energy is given by $\Delta U=mC_v\Delta T$, not $mC_p\Delta T$. If you divide by the heat capacity at constant volume rather than the heat capacity at constant temperature, you will get the correct temperature rise of 80C.

Chet

13. Nov 27, 2014

### Alettix

Oddly, in Swedish high school we are not thaught that there are different kinds of heat capacities. I have never encountered the terms "heat capacity at constant volume" and "heat capacity at constant pressure" (but I guess the latter is the one we use in school). I guess I have a couple of things to google now! :) After that I will hopefully be able to understand this.

Once again, Thank you for your answeres!

14. Nov 27, 2014

### Alettix

Thank you, I found it very useful! :)

Last edited by a moderator: May 7, 2017
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