Thermodynamics

roastdismalMechanics

Oct 27, 2013 (4 years and 12 days ago)

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Thermodynamics

The study of heat energy

through random systems

Laws of Thermodynamics


0th Law: If Ta=Tb and Tb=Tc then Ta=Tc



1st Law: ΔE=W+Q (Conservation of
Energy)


The increase in thermal energy of a system is
determined by the heat of the system and the
work added to it.


The idea behind the
Heat Engine

Efficiency



W
net
=Q
net
=Q
hot
-
Q
cold


or Eff=W
net
/Q
hot
=1
-
(Q
cold
/Q
hot
)



Ex: A steam engine absorbs 1.98 x10
5
J and
expels 1.49 x10
5
J in each cycle. Assume that all
the remaining energy is used to do work.


a. What is the engine’s efficiency?


b. How much work is done in each cycle?



a. Eff=0.247


b. W=4.9 x10
4
J




2nd Law: A system can be defined by its
entropy. In a closed system entropy
tends to increase.


Entropy: A measure of the disorder
(randomness) of the system


3rd Law: Temperature and Entropy are
absolute scales.


At some point no temperature or entropy
exists.


T

0 K and S

S
0

Heat Engines


Engines allow heat energy to be transformed
into work or mechanical energy.


Work or Energy

Heat increases


no big deal. Friction does this.


Heat

Work or Energy


is a big deal. Heat is easy to move around. You could just
bring heat wherever you needed work done and “Boom!”
you

wouldn’t have to do the work, a machine could.


Work or Energy

Heat decreases


is also a big deal. Making food cold preserves it and allows it
to be moved readily. Less spoilage means less disease.


Any heat engine works on the same
properties.



A
hot reservoir

is the source of the energy.



Both words mean something. Hot means that
there is plenty of heat energy and reservoir means
that if heat is removed the temperature doesn’t
drop much.



There is also a need for a
cold reservoir.

Again, both words mean something.



Cold because it is at a lower temperature than
the hot reservoir and reservoir because it must be
large enough that you can dump heat into it
without appreciably raising the temperature.

What happens if we put a hot and cold reservoir in
contact? Thermal Equilibrium is
not
the answer!



Heat transfer (or flow) is the answer.


Remember that these are reservoirs so it would
take a long time for them to come into thermal
equilibrium.


This is great, but we don’t get any work out
of it.


We need to “steal” some of the energy
leaving the hot reservoir and make it do work
for us.

Heat Engine


Stealing some energy to do work

High Temp

Low Temp

ΔE

Q

W

Types of Heat Engine


Steam Engine

Internal Combustion Engine

Unfortunately we don’t get an even
trade!


We lose energy to randomness/Entropy


This is the 2nd Law of Thermodynamics



Automobile engines are only about 15%
efficient. That means for every 100J of heat
energy, 15J worth of work is done on the
piston and 85J of heat are discarded. Still,
this is the source of energy for most of our
transportation.




If a steam engine takes in 2.254 x 10
4

kJ of
heat and gives up 1.915 x 10
4

kJ of heat to the
exhaust, what is the engines efficiency?


We can go backwards!


This is a refrigerator or air conditioner

High Temp

Low Temp

ΔE

Q

W

This requires Energy/Work


This is why your refrigerator must be
plugged in.


It is constantly dumping heat into your
kitchen


Due to the 2nd Law of Thermodynamics
more heat is dumped than is removed


If you left the refrigerator door open you
would heat up the kitchen

Perfect Heat Engine


Q
cold


Q
hot

Δ
E

W


Q
cold


Q
hot

Δ
E

W

Efficiency



W
net
=Q
net
=Q
hot
-
Q
cold


or Eff=W
net
/Q
hot
=(1
-
Q
cold
/Q
hot
)



Ex: A steam engine absorbs 1.98 x10
5
J and
expels 1.49 x10
5
J in each cycle. Assume that all
the remaining energy is used to do work.


a. What is the engine’s efficiency?


b. How much work is done in each cycle?



a. Eff=0.247


b. W=4.9 x10
4
J



More 2
nd

Law


Entropy







S=entropy


Q=Heat (Joules)


T=Temperature (In Kelvin)


Entropy in a system must increase or at
least stay the same!!!!!!!!!!!!


An engine has a hot reservoir at 1000 K
and uses the atmosphere at 300 K as the
cold reservoir. You take 2500 J from the
hot reservoir to do 1900 J of work.


A. How much heat goes into the
atmosphere?


B. Is this engine possible? (Does the entropy
increase?)


W
net
=Q
net
=Q
hot
-
Q
cold

Q
cold
=Q
hot
-
W
net
=2500J
-
1900J =600J



Entropy









Engine is not possible.


What is the maximum amount of work we can take out?


How much work is done?

Carnot Engine


Maximum efficiency


In theory could run backwards





All temperatures in Kelvin


What is the efficiency of an ideal steam
engine with steam at 685 K and exhaust at
298 K?


What is Q
cold
?

Carnot Engine


Maximum efficiency


In theory could run backwards





All temperatures in Kelvin


What is the efficiency of an ideal steam
engine with steam at 685 K and exhaust at
298 K?


What is Q
hot

if Q
cold

is 450J?