# Why is heat measured in joules?

1. Mar 20, 2014

### Entanglement

My limited knowledge tells me that joules= newton.meters " W = F.d How can this equation be related to heat since this equation depends on force done for a certain displacement which seems unrelated to heat, when something produces heat where is this such force and displacement ????

2. Mar 20, 2014

### A.T.

Heat is microscopic kinetic energy, stored in the movement of molecules. If you apply a net force F over a distance d to a mass m, it gains the kinetic energy: F * d = 1/2 * m * v2

3. Mar 20, 2014

### UltrafastPED

Heat is measured in joules because James Prescott Joule experimentally showed that mechanical energy is converted to heat in 1843.

See http://en.wikipedia.org/wiki/James_Prescott_Joule

For more details study thermodynamics and statistical mechanics.

4. Mar 20, 2014

### CWatters

Try playing with a bicycle tyre pump?

5. Mar 23, 2014

### Andrew Mason

Just to add what others have said, Joule's experiment involved applying a force through a distance to churn water in a tank and measuring the change in temperature. He found that the change in temperature of a tank of water was proportional to the work done. So it is quite natural for heat and work to be measured in Joules.

AM

6. Mar 24, 2014

### luuurey

This interpretation is a bit missleading. Microscopic kinetic energy(of molecul movements) corresponds with temperature rather then with heat. Heat doesn't equal temperature. Heat is simply energy that a system gains or gives up. Strictly speaking, heat is transfer of energy. Thus heat is measured in the same units as energy, i.e. joules.

7. Mar 24, 2014

### Staff: Mentor

The microscopic kinetic energy of the molecules does not correspond to temperature. The quantity that corresponds to temperature is $\frac{\partial{E}}{\partial{S}}$ where $S$ is the entropy; this is what determines which direction the heat is transferred between two bodies as they move towards equilibrium.

The change in the microscopic kinetic energy of the molecules is exactly equal to the amount of heat transferred, which is why (to answer OP's question) we measure heat in Joules. If I have two bodies, one warmer than the other, and I bring them into contact with each other, heat will flow from the warmer one to the cooler one. When they reach equilibrium, I will find that the total kinetic energy of the molecules in the one has increased by exactly the same number of Joules as the the total kinetic energy of the molecules in the other has decreased. Thus, it's very natural to think of the transfer of heat as the transfer of some number of Joules of energy.

8. Mar 24, 2014

### sophiecentaur

There is an interesting history to this. The original idea was that Heat was a substance (called Caloric) which would flow out of 'fire' and enter objects, as they got hotter. This idea was disproved by experiments which involved a lot of work, boring out a canon, and lots of heat. However, the start and end (total) mass turned out to be the same so it was concluded that heat was not an actual substance.

Even as recently as the 1960s, schools taught the 'Mechanical Equivalent of Heat', which is 4.2 Joules per Calorie, relating Work to Heat. (Same value as today but the name is not used any more.

9. Mar 24, 2014

### stevendaryl

Staff Emeritus
That's true, but for an ideal gas at constant volume, all the heat added to the gas goes into kinetic energy. If the volume is allowed to change, then heat added can go into expanding the volume in addition to increasing the kinetic energy.

10. Mar 24, 2014

### stevendaryl

Staff Emeritus
For an ideal gas, the internal energy $U$ is simply an accounting of the kinetic energy of the molecules. So a change in $U$ means a change in kinetic energy.

But not all heat added goes into changing $U$. By the equation

$dU = dQ - dW$

or rearranged:

$dQ = dU + dW$

we can see that only some of the heat added goes into kinetic energy, and some of it goes into performing work (expanding the volume against a pressure).

11. Mar 24, 2014

### luuurey

I can't agree with the opinion that kinetic energy of a particle of gas does not correspond to temperature. Each degree of freedom has energy kT/2. Mean kinetic energy of molecules (regardless of number of atoms) means 3 degrees of freedom. It means that the mean kinetics energy of molecules is 3kT/2. I don't know what your definition of the word 'to correspond' is, but it is obvious there's a relationship between mean kinetic energy of molecules and temperature. I guess, it was Feynman who said in his lectures that we could possibly measure temperature in joules instead of kelvins since temperature is nothing else than mean kinetic energy of the molecules.

However, everyone will hopefully agree that heat is just energy transferred from one object to another. We shouldn't confuse it with internal energy.

12. Mar 24, 2014

### Entanglement

Isn't the expansion of the gas as a result of weakening the intermolecular forces which results from the increase of K.E, it all happens due to the increase of K.E, what do you mean by a part increases the K.E while the other part make the gas to expand????

13. Mar 24, 2014

### stevendaryl

Staff Emeritus
When a gas expands, it cools down--the temperature goes down, and so does the average kinetic energy. So if you do the following:

1. Heat up a gas (this causes its average kinetic energy to go up)
2. Allow the gas to expand (this causes its average kinetic to go down)

So after these two steps, the net effect is to raise the temperature (and average kinetic energy) a little, and to increase the volume a little bit.

14. Mar 25, 2014

### nasu

No, the intermolecular forces are neglected in the ideal gas model.
And are negligible in real gases in "normal" conditions.
They play no part in the expansion of the gas.

15. Mar 25, 2014

### Entanglement

I thought increase in temp -------> increase in volume ------> increase in K.E how does the expansion let the K.E go down ??

Last edited: Mar 25, 2014
16. Mar 25, 2014

### Entanglement

Then how are gases expanded ?????

17. Mar 25, 2014

### sophiecentaur

If you let a gas expand against, say, a piston in a cylinder, the piston will be moving away, against some restraining force and the momentum change as the molecules strike the piston will result in their speeds being decreased slightly. The loss of the KE of the molecules is made up for by the gain of in energy of the piston and its restraint. This all refers to an ideal gas - no inter-molecular effects in the gas are involved.

18. Mar 25, 2014

### nasu

By increasing the volume available, for example. A gas will fill all the volume available to it. No external agent is necessary.
And this is because the inter-molecular forces are negligible in gases. The container is what keeps the gas from expanding and not inter-molecular forces. If the container cannot yield there will be no expansion, no matter how much you heat up the has and increase kinetic energy.

I don't really see how would think that expansion needs inter-molecular forces.
Maybe you have in mind a condensed phase, like liquids and solids. Here the thermal expansion is due not to the "weakening of intermolecular forces" but to an increase in thermal energy combined with the non-harmonicity of the intermolecular potential.

19. Mar 25, 2014

### stevendaryl

Staff Emeritus
Well, imagine a molecule hitting a wall. The molecule will bounce off the wall. If the wall is moveable, then some of the kinetic energy of the molecule will be transferred to the wall, causing the wall to move out, and therefore causing the volume to increase.

The energy $\delta E$ expended on pushing the wall outward is given by:

$\delta E = F \delta x$

where $\delta x$ is the distance the wall has been pushed out, and $F$ is the force exerted on the wall. In the case of a gas, the force on the wall is just $P \cdot A$, where $P$ is the gas pressure, and $A$ is the area of the wall. Putting these facts together gives:

$\delta E = P A \delta x = P \delta V$

The quantity $A \delta x$ is the change in volume of the box resulting from pushing the wall, of area $A$ out a distance $\delta x$.

So when the box expands, the energy transferred from the gas to the wall is given by:
$\delta E = P \delta V$. That is the energy that is taken away from the kinetic energy of the gas molecules.

20. Mar 26, 2014

### Entanglement

But are intermolecular forces involved in the contraction of gases in case of cooling ???