The fact that heat, Q , and work, W, can only determine internal energy, U , as inexact differentials, here represented by 'D', as in: dU = DQ + DW, (2-1)

which means that they can have any value and only their sum determines the internal energy and their actual value to reach a certain internal energy is path dependant, is sometimes quoted to say that thermodynamics can not provide a relationship between heat and work.

I like to provide another system which can use inexact differentials, and which may improve the understanding of internal energy: The volume of a vase.

I shall compare the volume of a trumpet shaped fresh clay vase with internal energy.

The volume of a fresh clay vase v can be altered by adding rings on the top and altering its height h or by changing the cross-sections of the vase as it exists. We call these changes Dv(height) and Dv(shape). dv = Dv(height) + Dv(shape).(2-2)

Back to thermodynamics: When I heat a system at constant volume, the heat required to increase the entropy S is dQ = T dS, where S is entropy. Earlier in the heating process the entropy increase required less heat since the temperature T was then lower. The heat content of the system heated at constant volume or no work addition is therefore:

Q = Integral(0 to S) T dS (2-3)where T changes from 0 to T

I shall now compare The equations in both systems.

Thermodynamics------------- Pottery

dU = DQ + DW (2-4)---------dv = Dv(height) + Dv(shape)(2-5)

DQ = T dS (2-6)-------------Dv(height = c dh (2-7)

DW = F dX (2-8)--------------Dv(shape) = Integral(0 to h) (dc) dh (2-9)

F = Integral(0 to S) (dT/dX) dS (2-10)

F dX = Integral(0 to S) (dT) dS (2-11)

Equation (2-6) states that heat addition to a system increases the entropy of the system. The heat requirement of that entropy addition is the temperature of the last entropy added.

Equation (2-11) states that the addition of mechanical energy to a system increases the heat requirements of the entropy differentials the system already has.

Thermodynamics claims that this relationship holds only for internal energy. It claims that external energy, energy due to displacement in external force fields, do not hold to this relationship. I disagree.

Equation (2-11) can not be evaluated, because 1. We do not know the absolute value of entropy and 2. the temperature change can only be calculated as long as the present phase of the system is stable. This means that only a small portion at the high end of the integral can be evaluated. To find out if a certain work addition applies to the relationship, we can mainly determine if the work addition results in a temperature rise at all or if that rise is comparable to the temperature rise due to the same work addition during compression.

From an SI steam table we can learn by double interpolation that the compression of a gram of steam at 1018.4 degree C from 101 to 100 cc results in a temperature rise of 2.63 degree, requiring a work input of 460 N cm/g. If the gram of steam were instead lifted against gravity by 1 cm, that requires .0098 N cm/g. We would expect that to result in a temperature rise of 2.63*0.0098/460 = 5.6*10 exp(-5) degree C or .56 degree C/ 100 m. It is likely that a small temperature change like that was not measured or, if measured was likely accepted as a measurement error, since classical thermodynamics assumes it to be zero.

Work of displacement in an electrical or magnetic field can be more easily created by altering the field strength experimentally. Such temperature changes are known as electro - or magneto - caloric effect