1. Chapter 4 Heat and the first law of thermodynamics
JIF 314 Thermodynamics
Chapter 4
Heat and the first law of thermodynamics

2. Key concepts Work and heat Adiabatic work
Work done adiabatically is path independence
Internal energy function, U is a state fuction of the system.
Wif (adiabatic) = Uf – Ui
Definition: Uf >Ui when work is done on the system.

3. Internal energy function
Interpretation of DU = Uf – Ui
Conservation of energy
Difference in energy function = energy change in the system
U a function of two thermodynamical coordinates, e.g. {P,T}, {P,V},{V,T}
The third variable is fixed by the equation of state.

4. U is specified by any two thermodynamical coordinates (out of three)
Example: Given the equation of state, PV = RT, there are two degree of freedom with three variables. Out of three variable, only two are independent since the equation of state fixes the third one.
For example, if we choose {V ,T } as the two independent variables, P is then the dependent variable that is fixed by the equation of state via
P =RT / V.
Alternatively, we can also choose {P ,T } as the two independent variables, V is then the dependent variable via the equation of state,V = RT /P .
As a conclusion, to specify the state of U, we need only any pair of independent thermodynamical coordinates

5. U is path-independent
U is path independent, hence dU is an exact differential
Say U (a, b), U (a+da, b+db), a,b any two thermodynamical coordinates
dU = U (a+da, b+db) - U (a, b)
{a, b} can be e.g. {T, V } or {T, P } or {V, P }.
In each case, the third variable, c, are P, V and T respectively.

6. are two different functions≠
U = U (a,c )
U = U (a,b)

7. Non-adiabatical process
Diathermal wall
For non-adiabatic process, Wif ≠ Uf – Ui
The difference between Wif and Uf – Ui is called heat, Q = (Uf – Ui) - Wif
Convention: Q is positive if heat enters a system, negative when if leave the system
Transit of heat is a result of a temperature difference
Heat is a form of energy

8. Q = (Uf – Ui) - W
Volume expand from Vi to Vf , hence work is done by the system. Internal energy increases from Ui to Uf when positive Q flows in. Work is done by the system.
weight
Volume expands
Temperature Tsu > Tsy so that heat flows from surrounding (su) to the system (sy)
U i
U i  U f
Diathermal wall permitting heat flow Q

9. Q and W have meaning only if a state undergoes transitional process to a new one
Heating and working are transient processes that causes a system to change from one state to another.
Heat and work are involved only in the process of making transition from a state to another.
Once the transition of states ceases and equilibrium achieved, heat or work exist no more.
Once the transition of state ceases, what endures finally is the new state, and the final internal energy.

10. Infinitesimal amount of Q, W are not exact differentials
U is a state function of the coordinates of the system, hence, it is path-independent
The difference in U between two infinitesimally different states is an exact differential, dU, and we can write, e.g.

11. Infinitesimal amount of Q, W are not exact differentials
In contrast, Q and W are not state function, path-dependent.
The difference in Q and W between two infinitesimally different states are not exact differential,
that is, e.g.,

12. Inexact differential form of Q and W
Hence, we use to denote an infinitesimal amount of heat, and not the differential form, dQ.
The same notation goes to W.

13. Calculation of W and Q are path-dependence
What all these mean are that: the calculation involving heat and work is path-dependent, and normally we have to carry out integration to determine W and Q. between two states, that is path-dependent.

14. Path-independence and path-dependence
As an example, when we calculate the difference in internal energy between two states, we only need to calculate the difference, DU =Uf – Ui. This difference is always the same since U is a state function. This infers path-independence.

15. However, in calculating the work done, DW when a system change from one state to another, we cannot simply calculate DW as Wf – Wi but we have to perform the integration which will result in different value for process carried out via different route (e.g. adiabatical path result in a final work done that is different from that is a non-adibatical one)

16. Within an adiabatic boundary, the heat lost (or gained) by system A is equal to the heat gained (or lost) by system B
Q = - Q’
Q, heat flom into system A
System A
Q’, heat flow into system B
System B
Adiabatic wall
diathermal wall

17. Differential form of the first law
Two inexact differentials on the right hand side (RHS) make one exact differential on the LHS.
For hydrostatic system (fluid),
and the first law reduces to
Work done is path dependent

18. Heat Capacity In unit of joules per kelvin (J/K)
It is a extensive quantity (i.e. the larger the mass the larger is the value of C since a larger amount of heat is require to heat up the material for 1 degree.)

19. Specific heat Capacity
In unit of joules per kelvin per kg (J/kg∙K)
Intensive quantity, i.e. it’s value remains the same for different amount of mass of the same material.

20. Molar heat capacity
n is the amount of material measured in unit of mole.
In unit of joules per kelvin per mole (J/mol∙K)
Intensive quantity, i.e. it’s value remains the same for different amount of mass of the same material.

21. Amount of material in mole
1 mole of material = NA atom
NA = Avogardo number,  1023
If an atom has a mass of m, N atoms will have a total mass of M = mN
Given an element with atomic mass m, M kg of material made up of such element contains
N = M/m atoms.
The ratio of N/NA defines the amount of atom in mole of that material: n = N/NA

22. Heat capacity at constant pressure

23. Heat capacity at constant volume

24. Equations for a hydrostatic system
If we choose U=U(T,V)

25. Special case, dV = 0 (for the case of CV)
Raising the temperature by heating substance but without changing the volume:
Specific heat at constant volume of a substances CV can be calculated from theory if the internal energy function of that substance is known.

26. Special case, dP = 0 (for the case of CP)
Raising the temperature by heating substance but without changing the pressure:
Specific heat at constant pressure of a substances CP can be calculated from theory if the internal energy function and b of that substance is known.

27. Heat reservior
A body of such a large mass that it may absorbed or reject an unlimited quantity of heat without experiencing an appreciable change in temperature or in any other thermodynamic coordinate.

28. Calculating quasi-static isobaric heat transfer process via a temperature difference
If CP is constant in temperature in the range of Ti – Tf,

29. Calculating quasi-static isochoric heat transfer process via a temperature difference
If CV is constant in temperature in the range of Ti – Tf,

30. Three mechanism of heat conduction
Convection
Radiation

31. Heat conduction Heat flow from high temperature to low temperature
Thermal conductivity
Cross section perpendicular to direction of heat flow
Temperature gradient

32. Heat convection
Convection coefficient
Temperature difference

33. Thermal radiation Emission of heat as electromagnetic radiation
Absorbitivity
Radiant exitance, R
Emissivity, e
Black body
Kirchhoff’s law
Radiated heat

34. Stefan-Boltzmann law
Stefan-Boltzmann constant, =  10-8 W/m2∙K4

35. Experimental determination of s
Nonequilibrium method
Equilibrium method

36. Question 4.1
Regarding the internal energy of a hydrostatic system to be a function of T and P, derive the following equations: a) b) c)

37. Solution for 4(a) U =U (T, P ) First law of Thermodynamics =>
Combining both
Eq. (1)

38. Solution for 4(a) (cont.)
For a hydrostatic system or PVT system, we can write V as a function of T and P.
By substituting the expression of dV into equation Eq(1), we get
Eq. (2)

39. Solution for 4(b)
At constant pressure, dP=0. Setting dP=0, and dividing Eq. (2) by dT, we get
Since
Therefore,

40. Solution for 4(c)
At constant volume, dV=0. Setting dV=0, and dividing Eq. (1) by dT, we get
Eq. (3)
Eq. (4)

41. Solution for 4(c)
Combining Eq. (3), (4), and

42. Question 4.3
One mole of a gas obeys the van der Waals equation of state:
and its molar internal energy is given by where a,b,c and R are constants. Calculate the molar heat capacities cv and cP.

43. Solutions We write U = U(T,V) Eq. (1) Eq. (2)
Eq. (1) combined with Eq. (2)
dT
Eq. (5)

44. Solutions
At constant volume, Eq. (5) becomes

45. Solutions
At constant volume, Eq. (5) becomes
From