Energy transfer by work and the first law

Feb 2020
4
0
Colombia
Why literature (Thermodynamics an Engineering Approach Eghth Edition by Cengel-Boles) say that heat and work are path functions and their magnitudes depends on the path followed (pg 63) and after when the first law is defined, the same book say. "For all adiabatic processes between two specified states of a closed system, the net work done is the same regardless of the nature of the closed system and the details of the process.", process = path? (pg 70) then W depends on path followed or not?, or the statement before is only for adiabatic processes? I am confused
 
Apr 2015
1,227
356
Somerset, England
OK so

The basic idea is that the net change of a state variable between two states - in this case the internal energy E or U - depends only upon the states, not the path.

So the change in internal energy going from state A to state B = \(\displaystyle {U_B} - {U_A}\)

The first law tells us that the internal energy is made up of a combination of heat transfer and work done.

Neither heat transferred nor work done are state variables.

(I would have to look up the book's sign convention for this)

But the definition of adiabatic is zero heat transfer.

So all energy changes are due to work done.

Note the definition also of a closed system is one with no mass transfer, but allows energy transfer.

Does this help?
 
Feb 2020
4
0
Colombia
ok but, work done between two states is the same regardless process or path?
 
Apr 2015
1,227
356
Somerset, England
No that is not OK.

Work done depends upon the path, in general.

In the special case of no other energy transfer (heat inpout) yes it is path independent.

Read through my chain of reasoning in my previous post again, and ask again if it is still not clear.

Note 'the process' and 'path' are two ways of looking at the same thing.
The 'states' are absolute values, which do not depend upon either process or path.
 
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Apr 2015
1,227
356
Somerset, England
Good stuff.

Cengel and Boles is a very good modern book.
Mine is the first Edition I think since it is dated 1989 and does not mention editions.

In my edition they state the first law as (page 99)

ΔE = Q - W

Now this is the Engineer's or Physicist's convention

Chemists would write this as

ΔE = Q + W

This is not a different equation or law. Just a different point of view.

The Engineer's version comes from the days of steam engines, where you put heat into and got work out of the steam.

So the change in internal energy of the steam was = the heat input minus the work output ( both considered positive.)

Chemists say

The change in internal energy is = the sum of all types of energy inputs with input considered positive and outputs considered negative.

So beware of sign conventions in other books.

Happy studying of Thermo.
 
Feb 2020
4
0
Colombia
Ok

(Qin - Qout) + (Win - Wout) = (Qin - Qout) - (Wout - Win)

But I didn't that the right side expression come from the days of steam engines
 
Apr 2015
1,227
356
Somerset, England
In the only steam engines I have ever seen you put heat into the water to make the steam, but no work.
Then the steam does work of expansion for you, but you do not put in or take out heat.

The thing to learn initially is that the heat and the work cross the system boundary in one direction or the other.
This is why they are not state variables.
They do not represent the state of the system.

State variables are quantities(properties) like temperature, pressure, volume, internal energy and so on.
These are properties of the system alone, not the surroundings.
The surroundings may have corresponding values of these same properties, independent of the system.


Work and heat transfer are variables of exchange between the system and its surroundings.