Download Chapter 15 Lesson 2

Chapter 15 Lesson 2
THE ZEROTH LAW OF THERMODYNAMICS
Two systems individually in
thermal equilibrium
with a third system are in
thermal equilibrium
with each other.
MOLAR HEAT CAPACITY
The molar heat capacity C is defined as
the heat per unit mole per Celsius degree.
Check with your instructor to
see if this more thorough
treatment of thermodynamic
processes is required.
SPECIFIC HEAT CAPACITY
Remember the definition of specific heat
capacity as the heat per unit mass
required to change the temperature?
Q
c
m t
For example, copper: c = 390 J/kgK
MOLAR SPECIFIC HEAT CAPACITY
The “mole” is a better reference for gases
than is the “kilogram.” Thus the molar
specific heat capacity is defined by:
C=
Q
n T
For example, a constant volume of oxygen
requires 21.1 J to raise the temperature of
one mole by one kelvin degree.
SPECIFIC HEAT CAPACITY
CONSTANT VOLUME
How much heat is required to
raise the temperature of 2 moles
of O2 from 0oC to 100oC?
Q = nCv T
Q = (2 mol)(21.1 J/mol K)(373 K - 273 K)
Q = +4220 J
SPECIFIC HEAT CAPACITY
CONSTANT VOLUME (Cont.)
Since the volume has not
changed, no work is done. The
entire 4220 J goes to increase
the internal energy, U.
Q = U = nCv T = 4220 J
U = nCv T
Thus, U is determined by the
change of temperature and the
specific heat at constant volume.
SPECIFIC HEAT CAPACITY
CONSTANT PRESSURE
We have just seen that 4220 J of
heat were needed at constant
volume. Suppose we want to also
do 1000 J of work at constant
pressure?
Q = U + W
Q = 4220 J + J
Q = 5220 J
Same
HEAT CAPACITY (Cont.)
Heat to raise temperature
of an ideal gas, U, is the
same for any process.
U = nCvT
For constant pressure
Q = U + W
nCpT = nCvT + P V
Cp > Cv

Cp
Cv
REMEMBER, FOR ANY PROCESS
INVOLVING AN IDEAL GAS:
PV = nRT
Q = U +  W
PAVA
TA
=
PBVB
TB
U = nCv T
The second law is a statement about the natural tendency of heat to
flow from hot to cold, whereas the first law deals with energy
Conservation and focuses on both heat and work.
THE SECOND LAW OF THERMODYNAMICS:
THE HEAT FLOW STATEMENT
Heat flows spontaneously from a substance at a higher
temperature to a substance at a lower temperature
and does not flow spontaneously in the reverse direction.
A heat engine is any device that uses heat to
perform work. It has three essential features.
1. Heat is supplied to the engine at a relatively
high temperature from a place called the hot
reservoir.
2. Part of the input heat is used to perform
work by the working substance of the engine.
3. The remainder of the input heat is rejected
to a place called the cold reservoir.
QH  magnitude of input heat
QC  magnitude of rejected heat
W  magnitude of the work done
The efficiency of a heat engine is defined as
the ratio of the work done to the input heat:
e
W
QH
If there are no other losses, then
QH  W  QC
e  1
QC
QH
Example: An Automobile Engine
An automobile engine has an efficiency of 22.0% and produces
2510 J of work. How much heat is rejected by the engine?
e
W
QH  W  QC
QH
QH 
W
e
QH  W  QC
QH 
W
QC  QH  W
1 
QC 
 W  W   1
e
e 
W
 1

 2510 J 
 1  8900 J
 0.220 
e
A reversible process is one in which both the system and the
environment can be returned to exactly the states they
were in before the process occurred.
CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT
OF THE SECOND LAW OF THERMODYNAMICS
No irreversible engine operating between two reservoirs at constant
Temperatures can have a greater efficiency than a reversible engine
operating between the same temperatures. Furthermore, all reversible
engines operating between the same temperatures have the same efficiency.
QC
QH
TC

TH
The Carnot engine is usefule as an idealized
model.
All of the heat input originates from a single
temperature, and all the rejected heat goes
into a cold reservoir at a single temperature.
Since the efficiency can only depend on
the reservoir temperatures, the ratio of
heats can only depend on those temperatures.
QC
QH
e  1
QC
QH
TC
 1
TH

TC
TH
Example 7 A Tropical Ocean as a Heat Engine
Water near the surface of a tropical ocean has a temperature of
298.2 K, whereas the water 700 meters beneath the surface has
a temperature of 280.2 K. It has been proposed that the warm
water be used as the hot reservoir and the cool water as the cold
reservoir of a heat engine. Find the maximum possible efficiency for
such and engine.
ecarnot
TC
 1
TH
ecarnot
TC
280.2 K
 1
 1
 0.060
TH
298.2 K
Refrigerators, air conditioners, and heat pumps are
devices that make heat flow from cold to hot.
This is called the refrigeration process.
The heat pump uses work to
make heat from the wintry
outdoors flow into the house.
Example 10 A Heat Pump
An ideal, or Carnot, heat pump is used to heat a house at 294 K.
How much work must the pump do to deliver 3350 J of heat into
the house on a day when the outdoor temperature is 273 K?
QC
QH
TC

TH
TC
QC  QH
TH
W  QH  QC
 TC
W  QH 1 
 TH



 TC
W  QH 1 
 TH
For heat
pump

 273 K 
  3350 J 1 
  240 J
 294 K 

Coefficien t of performanc e 
QH
W
The greater the ratio of Coefficient of Performance,
the better the performance of the heat pump or air
conditioner/refrigerator.
In general, irreversible processes cause us to lose some, but not
necessarily all, of the ability to do work. This partial loss can be
expressed in terms of a concept called entropy.
Carnot
QC
engine
QH
entropy
change
TC

TH
QC
TC

QH
TH
Q
S   
 T R
reversible
Entropy, like internal energy, is a function of the
state of the system.
Q
S   
 T R
Consider the entropy change of a Carnot engine.
The entropy of the hot reservoir decreases and the
entropy of the cold reservoir increases.
S  
QC
TC

QH
TH
0
Reversible processes do not alter the entropy of the universe.
Any irreversible process increases
the entropy of the universe.
S universe  0
THE SECOND LAW OF THERMODYNAMICS STATED
IN TERMS OF ENTROPY
The total entropy of the universe does not change
when a reversible process occurs and increases
when an irreversible process occurs.
Wunavailable  To Suniverse
THE THIRD LAW OF THERMODYNAMICS
It is not possible to lower the temperature
of any system to absolute zero in a finite number
of steps.