# How is work transferred to the system recognised?

For example, a potato initially at room temperature $25 \sideset{^{\circ}}{}{\mathrm{C}}$ is baked in an oven that is maintained at $200\sideset{^{\circ}}{}{\mathrm{C}}.$

I made potato as the system and the outer surface of the skin as the system boundary. While the oven and the air inside it is the surroundings.

There is a temperature difference between the skin and the air in the oven which is the driving force of heat transfer (temperature difference).

What about work? Is there transfer through work done?

Isn’t the oven working to produce the heat in the oven which is then transferred to the potato? But work is pressure multiplied by the change in volume. However there’s no change in volume of the potato. So does this mean no work is done?

In summary, how do I identify whether work is done to the system or not?

• Would you want to count "work" that is extremely small in proportion to the heat transfer?
– JMac
Commented Aug 8, 2018 at 10:49
• @user185692 check out the following link page 4, example 1. It is an example of the last paragraph of my answer. Hope this along with my answer helps.sfu.ca/~mbahrami/ENSC%20388/Notes/… Commented Aug 11, 2018 at 12:39

There definitely is work being done.

There is power being used to keep the oven hot at the mentioned temperature; power used to increase the temperature of the potato; and water vaporization. However, for your particular application, only the power used to increase the temperature of the potato is of interest. This is why you are given the formula using pressure and volume. Although it is true that the potato's volume changes little, the pressure inside the potato increases a lot. The temperature increase causes physical and molecular changes of the potato molecules.

While it may be (I haven’t researched it) that the potato “expands” a bit during the cooking process, I would think the amount of work to do this would be extremely small (as JMac pointed out) compared to the heat transfer to the potato that increases its internal energy.

Regarding your question “isn’t the oven working to produce the heat in the oven which is then transferred to the potato? But work is Pressure multiply by the change in volume. However there’s no change in volume of the potato. So does this mean no work is done?”

No, it does not mean no work is done. Not all work is pressure multiplied by volume change. That type of work is properly called “boundary work” (expanding the potato skin). But there are other forms of work. One is electrical work. When the oven’s heating element heats the air in the oven, it is a work transfer (at a rate of $i^2R$ ) from the electrical source to the heating element and then heat transfer from the heating element to the air.

The First Law of Thermodynamics says that changes to the internal energy of a thermodynamic system into two ways of energy transfers.

Work refers to forms of energy transfer, which can be accounted for in terms of changes in the macroscopic physical variables of the system, e.g. energy which goes into expanding the volume of a system against an external pressure, by say driving a piston-head out of a cylinder against an external force.

This is in distinction to heat energy carried into or out of the system in the form of transfers in the microscopic thermal motions of particles.

Any net increase in the internal energy U of a thermodynamic system must be fully accounted for, in terms of heat delta Q entering the system less work delta W done by the system:

dU = delta Q - delta W

The Roman letter d indicates that internal energy U is a property of the state of the system, so changes in the internal energy are exact differentials - they depend only on the original state and the final state, not on the path taken.

In contrast, the Greek delta Q or delta W in the equation reflects the fact that the heat transfer and the work transfer are not properties of the final state of the system.

Given only the initial state and the final state of the system, all one can say is what the total change in internal energy was, not how much of the energy went out as heat, and how much as work.

Well, it is not strictly true that the volume stays fixed. The potato actually gets dilated in the oven, and the change in volume means that a part of the heat transferred has been converted to work (rigorously speaking the work is not pressure by volume because the potato is a solid body, so you should instead consider the internal stresses, but the concept is the same).