1. PHYSICS 231 Lecture 31: Engines and fridges
Remco Zegers
Question hours: Thursday 12:00-13:00 & 17:15-18:15
Helproom
PHY 231

2. U=Q+W Metabolism Work done (negative) Heat transfer: Negative
body temperature < room temperature
Change in internal energy: Must be increased: Food!
PHY 231

3. U = Q + W t Metabolic rate
Metabolic rate: rate in which food and oxygen are
transformed into internal energy (to balance losses due
to heat loss and work done).
|W/t|
Body’s efficiency:
|U/t|
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4. Body’s efficiency U/t~oxygen use rate can be measured
W/t can be measured
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5. Types of processes A: Isovolumetric V=0 B: Adiabatic Q=0
C: Isothermal T=0
D: Isobaric P=0
PV/T=constant
First law of thermo-
Dynamics:
U=Q+W
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6. Isovolumetric processes (line A)
W=0 (area under the curve is zero)
U=Q (Use U=W+Q, with W=0)
In case of ideal gas:
U=3/2nRT
if P then T (PV/T=constant)
so U=negative Q=negative
(Heat is extracted from the gas)
if P then T (PV/T=constant)
so U=positive Q=positive
(Heat is added to the gas) p v
PHY 231

7. Adiabatic process (line B)
Q=0
No heat is added/extracted from the
system.
U=W (Use U=W+Q, with Q=0)
In case of ideal gas:
U=3/2nRT
if T
U=negative W=negative
(The gas has done work)
if T
U=positive W=positive
(Work is done on the gas) v
PHY 231

8. T=0 isothermal processes p The temperature is not changed
Q=-W (Use U=W+Q, with U=0)
if V
W=positive Q=negative
(Work is done on the gas
and energy extracted)
if V
W=negative Q=positive
(Work is done by the gas
and energy added) v
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9. P=0 isobaric process p Use U=W+Q In case of ideal gas:
W=-PV & U=3/2nRT
if V then T (PV/T=constant)
W: positive (work done on gas)
U: negative
Q: negative (heat extracted)
if V then T (PV/T=constant)
W: negative (work done by gas)
U: positive
Q: positive (heat added) v
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10. Cyclic processes The system returns to its original state. Therefore,
the internal energy must
be the same after completion
of the cycle (U=0)
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11. Cyclic Process, step by step 1
Process A-B.
Negative work is done on the gas:
(the gas is doing positive work).
W=-Area under P-V diagram
=-[(50-10)*10-3]*[( )*105]
-½[(50-10)*10-3]*[( )]*105= =
W= J
U=3/2nRT=3/2(PBVB-PAVA)=
= 1.5*[(1E+5)(50E-03)-(5E+5)(10E-03)]=0
The internal energy has not changed
U=Q+W so Q=U-W=12000 J: Heat that was added to the
system was used to do the work!
PHY 231

12. Cyclic process, step by step 2
Process B-C
W=-Area under P-V diagram
=-[(10-50)*10-3*( )*105]=
W=4000 J
Work was done on the gas
U=3/2nRT=3/2(PcVc-PbVb)=
=1.5[(1E+5)(10E-3)-(1E+5)(50E-3)]=-6000 J
The internal energy has decreased by 6000 J
U=Q+W so Q=U-W= J= J
10000 J of energy has been transferred out of the system.
PHY 231

13. Cyclic process, step by step 3
Process C-A
W=-Area under P-V diagram
W=0 J
No work was done on/by the gas.
U=3/2nRT=3/2(PcVc-PbVb)=
=1.5[(5E+5)(10E-3)-(1E+5)(10E-3)]=+6000 J
The internal energy has increased by 6000 J
U=Q+W so Q=U-W= J=6000 J
6000 J of energy has been transferred into the system.
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14. Summary of the process -AREA A-B B-C C-A Quantity Process Work(W)
Heat(Q) U
A-B J
12000 J
B-C
4000 J J
-6000
C-A
0 J
6000 J
6000
SUM
-8000 J
8000 J
-AREA
A-B
B-C
C-A
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15. What did we do? The gas performed net work (8000 J)
while heat was supplied (8000 J):
We have built an engine!
What if the process was done in
the reverse way?
Net work was performed on the
gas and heat extracted from the gas.
We have built a heat pump!
(A fridge)
PHY 231

16. More general engine turns water to steam W=|Qh|-|Qc| heat reservoir Th
efficiency: W/|Qh|
e=1-|Qh|/|Qc| Qh
work is done
the steam moves the piston
work
engine W Qc
The efficiency is determined
by how much of the heat you
supply to the engine is turned
into work instead of being lost
as waste.
the steam is condensed
cold reservoir Tc
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17. Reverse direction: the fridge
heat is expelled to outside
heat reservoir Th Qh
work is done
a piston compresses the coolant
work
engine W Qc
the fridge is cooled
cold reservoir Tc
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18. The 2nd law of thermodynamics
1st law: U=Q+W
In a cyclic process (U=0) Q=W: we cannot do more work
than the amount of energy (heat) that we put inside
2nd law: It is impossible to construct an engine that,
operating in a cycle produces no other effect than the
absorption of energy from a reservoir and the performance
of an equal amount of work: we cannot get 100% efficiency
What is the most efficient engine we can make
given a heat and a cold reservoir?
PHY 231

19. Carnot engine AB isothermal expansion BC adiabatic expansion W-, T-
W-, Q+
Thot
DA adiabatic compression
W-, T-
CD isothermal compression
Tcold
W+, Q-
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20. Carnot cycle inverse Carnot cycle Work done by engine: Weng
Weng=Qhot-Qcold
efficiency: 1-Tcold/Thot
A heat engine or a fridge!
By doing work we can
transport heat
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21. Next lecture
Entropy and examples
PHY 231