PHYSICS 231 Lecture 31: Engines and fridges. Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom. Metabolism. U=Q+W. Work done (negative). Heat transfer: Negative body temperature < room temperature. Change in internal energy: Must be increased: Food!. t. t.
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Question hours: Thursday 12:00-13:00 & 17:15-18:15
Work done (negative)
Heat transfer: Negative
body temperature < room temperature
Change in internal energy: Must be increased: Food!
U = Q + W
Metabolic rate: rate in which food and oxygen are
transformed into internal energy (to balance losses due
to heat loss and work done).
U/t~oxygen use rate
can be measured
W/t can be measured
A: Isovolumetric V=0
B: Adiabatic Q=0
C: Isothermal T=0
D: Isobaric P=0
First law of thermo-
The system returns to its
original state. Therefore,
the internal energy must
be the same after completion
of the cycle (U=0)
W=-12000 JCyclic Process, step by step 1
Negative work is done on the gas:
(the gas is doing positive work).
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!
Work was done on the gasCyclic process, step by step 2
The internal energy has decreased by 6000 J
U=Q+W so Q=U-W=-6000-4000 J=-10000 J
10000 J of energy has been transferred out of the system.
W=-Area under P-V diagram
No work was done on/by the gas.
The internal energy has increased by 6000 J
U=Q+W so Q=U-W=6000-0 J=6000 J
6000 J of energy has been transferred into the system.
the reverse way?
Net work was performed on the
gas and heat extracted from the gas.
We have built a heat pump!
(A fridge)What did we do?
The gas performed net work (8000 J)
while heat was supplied (8000 J):
We have built an engine!
turns water to steam
heat reservoir Th
work is done
the steam moves the piston
The efficiency is determined
by how much of the heat you
supply to the engine is turned
into work instead of being lost
the steam is condensed
heat is expelled to outside
heat reservoir Th
work is done
a piston compresses the coolant
the fridge is cooled
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?
AB isothermal expansion
BC adiabatic expansion
DA adiabatic compression
CD isothermal compressionCarnot engine
inverse Carnot cycle
Work done by engine: Weng
A heat engine or a fridge!
By doing work we can
Entropy and examples