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First Law of Thermodynamics
Physics 102Professor Lee
CarknerLecture 6
“of each the work shall become manifest, for the day shall declare it, because in fire it is revealed, and the work of each, what kind it is, the fire shall prove”
-- 1 Corinthians 3
PAL #5 Phase Change Heat needed to melt Frosty (ice at –5 C to water at 20 C)
Q2 = mL = (100)(33.5X104) = 3.35X107 J Total = 4.2917X107 J
What is Frosty’s final temperature if Santa removes 45 million joules? Since 45X106 – 42.917X106 = 2.083X106 J, Frosty will be colder
than –5 C Frosty is 9.97 degrees colder than his original –5 C or Tf = -
14.97 C
Energy We know that in mechanics energy is
conserved
In what ways can energy be expressed?
Heat can flow in or out
Work can be done on it or by it
The internal energy might change e.g. by changing the temperature
Work and Internal Energy
No heat can travel in or out
If weight is removed from the piston head, the remaining weight will rise
It must come from the internal energy of the gas
Work and Heat
The thermal reservoir can add or subtract heat from the system
What happens to the internal energy of the system as heat is applied or work is done?
Work, Heat and Internal Energy
If we add weight and do 6 J of work we either increase the internal energy by 6 J or produce 6 J of heat or some combination that adds up to 6
The First Law of Thermodynamics
This conservation of energy is called the First Law of Thermodynamics
U = Q - W
If work is done by the system W is positive, if work is done on the system W is negative
Heat flow in is +, heat flow out is -
PV Diagram
How much work is done if a gas expands and raises a piston?
Depends on:
The relationship between P and V can be complicated, but Work equals area under curve in PV diagram
The P-V Curve
If the volume decreases, work is done on the system and the work is negative
If the process is cyclical and returns to the same point by two different paths the area between the paths is equal to the work (and also equal to the heat)
Internal Energy and Temperature
If all energy is in the kinetic energy of the molecules
The total internal energy is the sum of the
kinetic energies of the all the molecules U = (3/2)NkT for N molecules
Ideal Gas Specific Heats
Q = mcT Instead of the mass we usually have the number
of moles and so use the molar specific heat (C) Q = nCT
However, when we add heat to a gas it may cause the gas to expand and the energy to go into work instead of temperature
Specific Heats
Molar specific heat at constant volume (CV) QV = nCVT QV = nCVT = U = (3/2)nRT
Molar specific heat at constant pressure (CP) QP = nCPT QP = nCPT = (3/2)nRT + nRT
CP > CV since some heat goes into work for CP
Types of Processes
We want to understand 5 basic types of thermodynamic processes
For each you should know:
Isobaric
In an isobaric process the pressure does not change
Can use heat capacity at constant pressure:
Since the area under the PV curve is a square: W=PV U = nCpT-PV
Today’s PAL Consider a cylinder with a volume
of 2 m3 filled with 1 mole of an ideal gas at a temperature of 300 K
If the gas is compressed to 1 m3 at constant pressure, what is the magnitude and sign of the work
Isochoric
In an isochoric process the volume does not change
W = 0 soU = Q
We can also relate to specific heat at constant volume: Q= nCVT
Adiabatic Adiabatic processes are ones in which no heat is
transferred
Q=0 so U = -W
We can also find relationships with the ratio of specific heats = CP/CV
For any adiabatic process PV = constant TV-1 = constant
Cyclical Process
A cyclical process returns to its initial state
Eint = 0 so Q=W There are many different ways to
produce a cyclical process