Why is heat measured in joules?

[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 ?

 

[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

 

[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.

 

[CWatters]

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 ?

Try playing with a bicycle tyre pump?

 

[Andrew Mason]

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 ?

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

 

[luuurey]

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

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.

 

[Nugatory]

This interpretation is a bit misleading. Microscopic kinetic energy(of molecule movements) corresponds with temperature rather then with heat.

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.

 

[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.

 

[stevendaryl]

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.

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.

 

[stevendaryl]

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.

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).

 

[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.

 

[Entanglement]

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.

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?

 

[stevendaryl]

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?

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.

 

[nasu]

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?

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.

 

[Entanglement]

When a gas expands, it cools down--the temperature goes down, and so does the average kinetic energy. So if you do the following:[*]Allow the gas to expand (this causes its average kinetic to go down)

[/LIST]
.

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

 

[Entanglement]

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.

Then how are gases expanded ?

 

[sophiecentaur]

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

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.

 

[nasu]

Then how are gases expanded ?

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.

 

[stevendaryl]

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

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.

 

[Entanglement]

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.

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

 

[Entanglement]

I got your point guys, on heating a gas the molecules will gain K.E, if the walls of the container are movable a part of the force will be consumed in doing work to move these walls so the volume occupied by the gas increases, but if the walls aren't movable the volume will never expand and all the heat will just increase the K.E. But my question is what is the net change if the walls are movable, will the K.E increase, decrease, or stay constant?

 

[sophiecentaur]

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

I got your point guys, on heating a gas the molecules will gain K.E, if the walls of the container are movable a part of the force will be consumed in doing work to move these walls so the volume occupied by the gas increases, but if the walls aren't movable the volume will never expand and all the heat will just increase the K.E. But my question is what is the net change if the walls are movable, will the K.E increase, decrease, or stay constant?

The intermolecular forces are relevant when the molecules are close enough together. In the limit, this results in a change of state (condensation), during which the temperature is unchanged and all the energy change is taken up with the energy of the molecular attractions. (But this is not in the region where a gas can be considered as 'ideal').

You seem to want to discuss the effects entirely in words, when the Gas Laws describe it all, perfectly in a line of Maths.

P₁V₁/T₁ = P₂V₂/T₂

answers your question. You just plug in the values you want to change and the values you want to keep constant and it tells you what you want to know. The areas in the PV diagram show you the work done on or by the gas and energy is conserved. It is possible to add energy without increasing the temperature at all (isothermal), in which case, all the added energy is used up as work done on the moving cylinder.

 

[Entanglement]

The intermolecular forces are relevant when the molecules are close enough together. In the limit, this results in a change of state (condensation), during which the temperature is unchanged and all the energy change is taken up with the energy of the molecular attractions. (But this is not in the region where a gas can be considered as 'ideal').
You seem to want to discuss the effects entirely in words, when the Gas Laws describe it all, perfectly in a line of Maths.
P₁V₁/T₁ = P₂V₂/T₂
answers your question. You just plug in the values you want to change and the values you want to keep constant and it tells you what you want to know. The areas in the PV diagram show you the work done on or by the gas and energy is conserved. It is possible to add energy without increasing the temperature at all (isothermal), in which case, all the added energy is used up as work done on the moving cylinder.

I don't get how is the isothermal process done ?
I

 

[sophiecentaur]

I don't get how is the isothermal process done ?
I

Compress (or allow to expand) a gas in a situation where the temperature cannot change and you get Boyle's Law behaviour. PV is constant.
I think you should read about the Gas Laws, starting at the very beginning. Many of your questions indicate that you are not totally familiar with the basics of this stuff. These basics take up several weeks worth of lessons, usually.

 

[Entanglement]

I think you should read about the Gas Laws, starting at the very beginning. Many of your questions indicate that you are not totally familiar with the basics of this stuff. These basics take up several weeks worth of lessons, usually.

Thermodynamics was excluded from our syllabus last year due to the political situation in my country, in our syllabus this year there is a chapter about cryogenics, so I'm studying low temperature physics with very little thermodynamics knowledge, I know that is stupid, but that's not my fault, the situation in my country is horrible, but I'll try to do me best.

 

[Andrew Mason]

I don't get how is the isothermal process done ?
I

If T is constant and V increases then P has to decrease. If P decreases, the walls do not expand if the external pressure remains the same. So if T is constant, the external pressure has to decrease if there is an expansion.

If you take your example, if heat flows into a gas, T increases and P increases if V is constant. However, if the external pressure is constant, then an increase in V continues until P is equal to the external pressure. So this describes a constant pressure expansion.

AM

 

[Entanglement]

If T is constant and V increases then P has to decrease. If P decreases, the walls do not expand if the external pressure remains the same. So if T is constant, the external pressure has to decrease if there is an expansion.

If you take your example, if heat flows into a gas, T increases and P increases if V is constant. However, if the external pressure is constant, then an increase in V continues until P is equal to the external pressure. So this describes a constant pressure expansion.

AM

...

 

[Entanglement]

I think I'll have to study the gas laws first, let's start with Boyle's law

It states that pressure is inversely proportional to volume, is that because the no of collisions of the molecules with the walls of the container due to Brownian motion increase in a smaller volume ?

 

[jbriggs444]

It states that pressure is inversely proportional to volume, is that because the no of collisions of the molecules with the walls of the container due to Brownian motion increase in a smaller volume ?

It's not that the motion of each molecule increases. If temperature is held contstant, the average velocity of the molecules will not change. It is that there are more molecules. So more molecules hitting a given area of wall surface per unit time.

 

[Entanglement]

It's not that the motion of each molecule increases. If temperature is held contstant, the average velocity of the molecules will not change. It is that there are more molecules. So more molecules hitting a given area of wall surface per unit time.

Yeah I got that, It just increases the chance of its collision with the walls, we can say it changes the no of collisions per unit area and per unit time but the average kinetic energy doesn't change since the Temperature is constant

 

[Entanglement]

when we are talking about the pressure of gases, what is really meant, its pressure due to gravity?? or its pressure due to its K.E and its collision with the surroundings ?

 

[Entanglement]

Compress (or allow to expand) a gas in a situation where the temperature cannot change and you get Boyle's Law behaviour. PV is constant.

.

If I compressed a gas in a container of movable walls which isn't thermally isolated from the surrounding, that is considered as a isothermal process ??

 

[sophiecentaur]

If I compressed a gas in a container of movable walls which isn't thermally isolated from the surrounding, that is considered as a isothermal process ??

It would be far more efficient for you to read through and follow what a textbook has to say about all this. Look at This link and also Wikipedia will tell you all you need to know. Q and A is a very poor way of learning basic stuff, compared with going with the flow of the argument presented in a textbook - that has been tried and tested. Q and A is a bit like Brownian Motion; every so often you may randomly go in the right direction but most of the time you can end up off-target.

 

[stevendaryl]

If I compressed a gas in a container of movable walls which isn't thermally isolated from the surrounding, that is considered as a isothermal process ??

Isothermal means that it is kept at a constant temperature. One way to do this is to have a very large reservoir (or air, or water, or whatever) that is kept at a constant temperature, and do the compression so slowly that the gas in the container is always allowed to reach equilibrium with the reservoir.

 

[Entanglement]

Isothermal means that it is kept at a constant temperature. One way to do this is to have a very large reservoir (or air, or water, or whatever) that is kept at a constant temperature, and do the compression so slowly that the gas in the container is always allowed to reach equilibrium with the reservoir.

So during the slow compress the gas keeps gaining and losing heat to the surrounding medium at the same rate staying at an equilibrium position with surrounding, so its temperature stays constant

 

[stevendaryl]

So during the slow compress the gas keeps gaining and losing heat to the surrounding medium at the same rate staying at an equilibrium position with surrounding, so its temperature stays constant

Normally, compressing a gas would cause it to get hotter. But you can view isothermal compression as a limiting case of the following process:

Compress the gas a tiny bit.
Let it cool off.
Compress it a tiny bit more.
Let it cool off.
Etc.

The resulting pressure when you reach your final volume will be less than it would have been if you hadn't allowed it to cool off.

Isothermal compression results in this relationship between pressure and volume:

P \propto V^{-1}

If the gas were prevented from exchanging heat with the reservoir, the relationship would be:

P \propto V^{-5/3} (for a monatomic gas).

 

[sophiecentaur]

So during the slow compress the gas keeps gaining and losing heat to the surrounding medium at the same rate staying at an equilibrium position with surrounding, so its temperature stays constant

If you change the volume 'slowly enough' the temperature can remain 'near enough' constant. It's an ideal scenario. The other extreme case - adiabatic volume change, assumes that no heat enters of leaves but that again is only an ideal situation. Real life is always somewhere in between.

 

[Entanglement]

If you change the volume 'slowly enough' the temperature can remain 'near enough' constant. .

If it's not done slowly it will gain heat, but eventually it will lose it reaching to an equilibrium state with the surrounding medium, so the final result is the same whether it's done slowly or fast, why should be so careful about that point ??

 

[WannabeNewton]

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...

For a classical gas the equipartition theorem states that the kinetic energy of each mole of the gas is directly proportional to the temperature of the gas. The definition $T = \frac{\partial \bar{E}}{\partial S}$, or even better $S = -\frac{\partial F}{\partial T}$ in the Helmholtz representation, does not contradict this as the equipartition theorem is proven using the statistical equivalent of that latter definition i.e. $\bar{E} = -\frac{\partial}{\partial \beta}\ln Z$. Sure the kinetic energy is itself not temperature but for classical gases it is directly proportional to it as are the energies of other quadratic degrees of freedom.

To the OP: heat $\delta Q$ measures the extent to which the change $\Delta U$ in the internal energy of a system through a process fails to be adiathermal. As others have noted, one can perform simple experiments to show that a quantity $\delta Q$ must exist to account for the change in internal energy of a system between equilibrium states that is not already taken into account by the work done on the system during the process and we know such a quantity must exist simply due to the existence of processes which are the entire opposite of adiabatic and which involve no work done at all but still have a change in internal energy (just take a beaker of water and put a lit Bunsen-burner under it). Clearly then $\delta Q$ must be measured in Joules.

 

[stevendaryl]

If it's not done slowly it will gain heat, but eventually it will lose it reaching to an equilibrium state with the surrounding medium, so the final result is the same whether it's done slowly or fast, why should be so careful about that point ??

Well, if the temperature is (approximately) constant, then the process is (approximately) isothermal. It's just a matter of definition.

You're right that for an ideal gas, the final state doesn't depend on whether the process was isothermal, or not. But the total work done and the change in the entropy of whatever system was compressing the gas depends on whether it was isothermal or not.

 

[Jobrag]

During his honeymoon in Switzerland Joule along with Thompson, (history does not relate why Joules and Thompson happened to meet in Switzerland during Joule's honeymoon), attempted to measure the temperature difference between the top and bottom of a waterfall, history also fails to mention what Joule's bride Amellia thought of it.

 

[stevendaryl]

During his honeymoon in Switzerland Joule along with Thompson, (history does not relate why Joules and Thompson happened to meet in Switzerland during Joule's honeymoon), attempted to measure the temperature difference between the top and bottom of a waterfall, history also fails to mention what Joule's bride Amellia thought of it.

That's funny. Maybe the story was twisted for the sake of New Yorkers like me, but I thought t was Niagara Falls.

 

[Entanglement]

Well, if the temperature is (approximately) constant, then the process is (approximately) isothermal. It's just a matter of definition.
You're right that for an ideal gas, the final state doesn't depend on whether the process was isothermal, or not. But the total work done and the change in the entropy of whatever system was compressing the gas depends on whether it was isothermal or not.

In case of the isothermal process, If the walls of the container are movable, and then the gas is compressed, part of the energy will be used up to expand the volume and the other part will increase the gas's temperature which will eventually get lost to the surrounding reaching a state of thermal equilibrium ( the temperature won't change ) ?

 

[stevendaryl]

In case of the isothermal process, If the walls of the container are movable, and then the gas is compressed, part of the energy will be used up to expand the volume and the other part will increase the gas's temperature which will eventually get lost to the surrounding reaching a state of thermal equilibrium ( the temperature won't change ) ?

If it's isothermal, then none of the energy goes into raising the temperature of the gas.

Some of the energy required to compress it will go towards decreasing the volume, and some of the energy will go towards heating up the reservoir (by assumption, the reservoir is so large that this extra heat causes no significant change to the temperature of the reservoir.

 

[sophiecentaur]

That's funny. Maybe the story was twisted for the sake of New Yorkers like me, but I thought t was Niagara Falls.

Switzerland, almost certainly. I'm not sure Niagra is high enough, in any case. I seem to remember doing the sums and we're talking in terms of something like 0.1C, iirc.

I seem to remember another story about Joule who took enough food with him to provide himself with enough energy to climb a mountain (just enough to achieve the 'work done'). He forgot about the food needed just to keep warm and got back in a bit of a state. Hard to believe an experienced experimenter would make such a trivial mistake - but, brain the size of a planet, and all that.