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First Law of Thermodynamics
Physics 102
Professor Lee Carkner
Lecture 5
(Session: 104884)
PAL #5 Phase Change
Final temperature of melted Frosty
Four heats:
Warm up Frosty to 0 C: micecice(0-(-5))
Melt Frosty: miceLice
Warm up melted Frosty: mwatercwater(Tf-0)
Cool down air: maircair(Tf-20)
(100)(2100)(5)+(100)(333000)+(100)(4186)(Tf)+(9700)
(837)(Tf-20) = 0
1.05X106+3.33X107+4.19X105Tf+8.12X106Tf-1.62X108
=0
8.54X106Tf = 1.28X108
Tf = 15 C
Energy
We know that in mechanics energy is
conserved
In what ways can energy be expressed?
Heat can flow in or out
Related to expansion or compression
The internal energy might change
Related to temperature
Consider a piston of gas with
weight on the top and a thermal
reservoir at the bottom
Weight can be added or subtracted
so that the system does work on the
weight or the weight does work on
the system
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
DU = Q - W
If work is done by the system W is positive, if
work is done on the system W is negative
Positive work is the useful work we get out
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:
Pressure
Change in volume
The relationship between P and DV 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)
P-V Diagrams
Internal Energy and Temperature
Internal energy (U) is directly related to
temperature
High T, large U
Low T, small U
KE = (3/2)kT for one molecule
U = (3/2)nRT for n moles
Note that everything has some internal
energy, we want to know about the change
in internal energy (DU)
Types of Processes
We want to understand 5 basic types of
thermodynamic processes
For each you should know:
PV diagram
Isobaric
In an isobaric process the
pressure does not change
Since the area under the
PV curve is a rectangle:
W=PDV
Isochoric
e.g. a sealed hollow cylinder
W = 0 so DU = Q
If any heat is applied to the
system it goes directly into
internal energy
Isothermal
An isothermal process
happens at constant
temperature
Since DT = 0:
DU = 0 so Q=W
We can use calculus to
find the area under
the curve
W = nRTln(Vf/Vi)
Adiabatic
Adiabatic processes are ones in which no heat
is transferred
Since there is no heat:
Q=0 so DU = -W
P-V Diagram
Isobaric (P=const.)
P
Isothermal (T=const)
Adiabatic (Q=0)
Isochoric (V=const)
V
Cyclical Process
A cyclical process returns to its initial
state
DU = 0 so Q=W
There are many different ways to
produce a cyclical process
Next Time
Read: 15.4-15.6
Homework : Ch 15: P 7, 10, 17, 29
Test 1 next Friday
About 10 multiple choice (~25%)
About 4 problems (~75%)
Equation and constant sheet given
I have posted equation sheet and practice
problems
As a pot of water boils, the temperature of the
water,
A)
B)
C)
D)
E)
Increases
Decreases
Stays the same
Fluctuates unpredictably
It depends on the temperature of the stove
Water condenses out of the air onto a cold piece
of metal. Due to this condensation, the
temperature of the air around the metal,
A)
B)
C)
D)
E)
Increases
Decreases
Stays the same
Fluctuates unpredictably
It depends on the temperature of the metal
The temperature of a solid is held constant and
the pressure is lowered. When the pressure
gets very close to zero the solid will,
A)
B)
C)
D)
E)
Become a gas
Become a liquid
Stay a solid
Be at the triple point
Be at the critical point