Why doesn't Q=0 instead of Q=W if T=constant in first law?

[Henrybar]

My understanding of heat is the energy transferred as a result of temperature imbalance between systems.
If two systems at different temperatures are in contact with each other, a temperature change for both systems and an energy transfer Q is occurring.

It is known that Q=W for a closed system when T=constant according to the first law of themo, but how can there be heat (energy transfer as a result of temperature imbalance between systems) and yet no change in temperature of the system?

 

[Delta2]

When we have two bodies A and B and flow of heat Q say from A to B (that means that B is in lower temperature than A) then normally what happens is that this heat energy goes as whole as an increase in the internal energy of body B hence it increases its temperature and eventually (after some flow of heat for some time $\Delta t$ ) the temperature of B becomes equal to temperature of A.
However what happens during say an isothermal expansion of a gas, the energy heat Q that flows to the gas does not go to increase its internal energy, rather it goes as whole to do mechanical work W. So its not the internal energy (and hence not the temperature ) that increases, it is rather the work done by the system that increases. In other words all the heat energy that flows into the gas, is being transformed into mechanical work W, and nothing at all goes to internal energy of the gas, hence its temperature stays constant.
To state it again in a final simplified way:
When all or part of heat energy Q is being transformed to internal energy then the temperature increases.
BUT when all heat energy Q is being transformed to mechanical work, then it is the mechanical work W done by the gas that increases, while its internal energy and temperature remain constant.

 

[Chestermiller]

My understanding of heat is the energy transferred as a result of temperature imbalance between systems.
If two systems at different temperatures are in contact with each other, a temperature change for both systems and an energy transfer Q is occurring.

It is known that Q=W for a closed system when T=constant according to the first law of themo, but how can there be heat (energy transfer as a result of temperature imbalance between systems) and yet no change in temperature of the system?

Are you familiar with the first law of thermodynamics. If so, please write it out for us in equation form.

 

[FactChecker]

My understanding of heat is the energy transferred as a result of temperature imbalance between systems.

This is wrong. Heat is energy, whether it is transferred or not. It can be transferred, but that is not necessary. (I have been corrected on this.)

If two systems at different temperatures are in contact with each other, a temperature change for both systems and an energy transfer Q is occurring.

It is known that Q=W for a closed system when T=constant according to the first law of themo, but how can there be heat (energy transfer as a result of temperature imbalance between systems) and yet no change in temperature of the system?

If you are talking about the transfer of heat within one isolated system, rather than the transfer of heat from one system to another, then you can have Q=W=0.

 

[Chestermiller]

This is wrong. Heat is energy, whether it is transferred or not. It can be transferred, but that is not necessary.

In thermodynamics, heat is defined as energy in transit across the boundary of a system as the result of a temperature gradient at the boundary, or as a result of a temperature difference between the system and its surroundings.

 

[FactChecker]

In thermodynamics, heat is defined as energy in transit across the boundary of a system as the result of a temperature gradient at the boundary, or as a result of a temperature difference between the system and its surroundings.

Ok. I stand corrected (and surprised)

 

[FactChecker]

Ok. I stand corrected (and surprised)

I should not have been surprised since temperature is used to measure the average kinetic energy of the particles in the system. Heat is something else.

 

[Henrybar]

Are you familiar with the first law of thermodynamics. If so, please write it out for us in equation form.

Q=W-ΔU
Heat added to system=work done by system - change in internal energy of system

 

[Chestermiller]

Q=W-ΔU
Heat added to system=work done by system - change in internal energy of system

Wrong. Try again.

 

[Henrybar]

When we have two bodies A and B and flow of heat Q say from A to B (that means that B is in lower temperature than A) then normally what happens is that this heat energy goes as whole as an increase in the internal energy of body B hence it increases its temperature and eventually (after some flow of heat for some time $\Delta t$ ) the temperature of B becomes equal to temperature of A.
However what happens during say an isothermal expansion of a gas, the energy heat Q that flows to the gas does not go to increase its internal energy, rather it goes as whole to do mechanical work W. So its not the internal energy (and hence not the temperature ) that increases, it is rather the work done by the system that increases. In other words all the heat energy that flows into the gas, is being transformed into mechanical work W, and nothing at all goes to internal energy of the gas, hence its temperature stays constant.
To state it again in a final simplified way:
When all or part of heat energy Q is being transformed to internal energy then the temperature increases.
BUT when all heat energy Q is being transformed to mechanical work, then it is the mechanical work W done by the gas that increases, while its internal energy and temperature remain constant.

I should have mentioned my question does particularly concern the isothermal expansion of gas. I have a few things I would like final clarification on.
How is an isothermal expansion of a gas maintained if there is a temperature imbalance (and heat transfer)? Is it because the experimenter is not allowing the temperature to change? Or is because the energy simply goes into work only, but why would it only go to work?

 

[Henrybar]

Wrong. Try again.

Q=W+ΔU
Heat added to system=work done by system + change in internal energy of system

 

[Chestermiller]

Isothermal expansion does not mean that the temperature of the gas is constant throughout its volume during the entire expansion. It means that the gas is kept in contact with a constant temperature reservoir that is held at the original temperature of the gas as the gas expands. As the gas expands, its interior cools (temporarily) as a result of the work it does, and heat then begins to flow between the constant temperature reservoir and the gas. When the gas approaches its final thermodynamic equilibrium state, it again equilibrates with the temperature of the reservoir. So, in an "isothermal expansion," only the surface of the gas in contact with the reservoir is at constant temperature throughout the process. The rest of the gas experiences lower temperatures (until the end).