Thermodynamics
Chapter 15
Calculating Work
Work = area under Pressure vs.
Volume graph
W = Fd
F = PA
W=PAd
W = PDV
Calculus link
V2
W = - p dV
V1
Isochoric (isovolumetric)
•
•
No change in volume
W=0
Isobaric Process
• No change in pressure
• Volume expands and does work
• WORK = - AREA
Pressure vs. Volume of an Ideal Gas
Pressure (X105 N/m2)
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
Volume (mL)
25
30
35
40
Work from Graph: Example 1
A gas expands at a constant pressure of 1 X 105
N/m2 (~100 kPa, ~ 1 atm) from a volume of 10
mL to a volume of 35 mL. Calculate the work
done by the gas.
Pressure vs. Volume of an Ideal Gas
Pressure (X105 N/m2)
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
Volume (mL)
25
30
35
40
W = PDV
W = (1 X 105 N/m2)(0.035 L – 0.010 L)
W = 2500 J
Work from Graph: Example 2
A gas is cooled to produce a decrease in pressure.
The volume is held constant at 10.0 m3, and the
pressure decreases from 1 X 105 N/m2 to 0.5 X
105 N/m2. Calculate the work done on the gas.
Pressure vs. Volume
Pressure (N/m2)
1.20E+05
1.00E+05
8.00E+04
6.00E+04
4.00E+04
2.00E+04
0.00E+00
0
2
4
6
Volume (m3)
8
10
12
W = PDV
W=PX0
W = 0 J (No work is done)
Work from Graph: Example 3
A gas expands and increases in pressure as shown
in the next graph. Calculate the work done on
the gas. (Remember to include the entire area
under the graph).
Pressure vs. Volume
Pressure (N/m2)
1.20E+05
1.00E+05
8.00E+04
6.00E+04
4.00E+04
2.00E+04
0.00E+00
0
10
20
30
40
Volume (m3)
50
60
70
Pressure vs. Volume
Pressure (N/m2)
1.20E+05
1.00E+05
8.00E+04
6.00E+04
4.00E+04
2.00E+04
0.00E+00
0
10
20
30
40
Volume (m3)
ANS:
50
60
70
Example 4
Calculate the work done from 500 to 1000 cm3 on
the following graph. Remember to go all the
way down to the x-axis.
Isothermal Systems
• Area under the curve
W = - p dV = - nRT dV = - nRT dV
V
V
W = -nRT ln Vf = -piVi ln Vf = pfVf ln Vf
Vi
Vi
Vi
Isothermal Example
A cylinder contains 7.0 grams of nitrogen (N2).
a. Calculate the moles of N2
b. Calculate the work that must be done to
compress the gas at a constant temperature of
80oC until the volume is halved.
Pressure vs. Volume of an Ideal Gas
Pressure (N/m2)
1.20E+05
1.00E+05
8.00E+04
6.00E+04
4.00E+04
2.00E+04
0.00E+00
0.00E+00
1.00E-01
2.00E-01
3.00E-01
Volume (m3)
Top Line = at 500K
Bottom Line = at 300K
4.00E-01
5.00E-01
6.00E-01
Work and Cyclic Processes
• Cyclic process – gas returns to its original state
• Important for studying
– Steam engine
– Car engine
DU = 0
Q=W
• Work = Area enclosed
Pressure (N/m2)
Pressure vs. Volume
4.50E+05
4.00E+05
3.50E+05
3.00E+05
2.50E+05
2.00E+05
1.50E+05
1.00E+05
5.00E+04
0.00E+00
0
10
20
30
Volume (m3)
40
50
Work and Cyclic Processes
Calculate the work shown in the previous graph.
Heat (Q)
1. Heat – Energy transferred from one body to
another because of a difference in temperature
2. Cooking a turkey
hot oven  cooler turkey
3. Extensive Property – depends on amount of
material (iceberg vs. water)
4. Unit
•
•
Joules.
Calorie
calories
1. calorie – amount of heat energy needed to
raise the temperature of 1 gram of water by 1
degree Celsius (or Kelvin)
2. Not a nutritional Calorie
1 nutritional Calorie = 1000 calories (or 1 kilocalorie)
Converting between heat units
1 cal = 4.18 Joules
252 cal = 1 BTU
1054 Joules = 1 BTU
James Prescott Joule
•
•
•
•
Weight moves paddles
Friction from paddles warms water
Workfalling = Heatpaddles
Mechanical Equivalent of Heat
Heat: Example 1
How high would you have to climb to work off a 500
Calorie ( 500,000 cal) ice cream? Assume you mass 60
kg.
(500,000 cal)(4.186J/cal) = 2.09 X 106 J
Heat = Work
Heat = mgh
h = Heat/mg
h = 2.09 X 106 J/(60 kg)(9.8 m/s2) = 3600 m
h ~ 11,000 ft
Heat: Example 2
A 3.0 gram bullet travels at 400 m/s through a tree.
After passing through the tree, the bullet is now
only going 200 m/s. How much heat was
transferred to the tree?
KE = ½ mv2
KE = ½ mv2
KE = ½ (0.003 kg)(400 m/s)2 KE = ½ (0.003 kg)(200 m/s)2
KE = 240 J
KE = 60 J
Heat loss = 180 J to the tree
The Laws of Thermodynamics
1st Law
– Energy is conserved
DE = W + Q
2nd Law
– Natural processes tend to move toward a
state of greater disorder
– Heat goes hot to cold (Clausius)
– No device converts all heat to work (KelvinPlanck)
DS >0
The Laws of Thermodynamics
3rd Law
– The entropy of a pure crystal at absolute
zero is zero
DG = DH - TDS
Important Definitions
Thermodynamics – Study of the transfer of energy as heat
and work
System – Objects we are studying
Isolated System
Closed System
Open System
No mass or
energy may
leave/enter
Only energy may Mass and Energy
leave/enter, not
may leave/enter
mass (Earth)
First Law Sign Conventions
Heat is added to the system
Heat is lost
Work on the system
Work done by the system
+
+
-
First Law Example 1
2500 J of heat is added to a system. This heat does
1800 J of work on the system. Calculate the
change in internal energy.
DE = W + Q
DE = 1800 J +2500 J
DE = 4300 J
First Law Example 2
2500 J of heat is added to a system. This increase
in temperature allows the system to do 1800 J of
work. Calculate the change in internal energy.
DE = W + Q
DE = -1800 J +2500 J
DE = 700 J
Adiabatic Systems
– No heat exchange (Q=0)
– Fast processes
– Heat has no time to enter/leave system
– Car piston
Q= 0
For an ideal gas
DE = W + Q
DE = 3/2 nRT
DE = W
DE = 3nRDT
2
Work: Example 1
In an engine, 0.25 moles of gas in the cylinder
expand adiabatically against the piston. The
temperature drops from 1150 K to 400K. How
much work does the gas do? (the pressure is not
constant).
Q= 0
DE = Q-W
DE = W
DE = 3nRDT
2
DE = 3(0.25 mole)(8.315 J/K-mol)(400 K - 1150 K)
2
DU = 2300 J
W = 2300 J
Temperature and Internal Energy
Temperature – measure of the average kinetic
energy of individual molecules
Internal (Thermal) Energy – Total energy of all the
molecules in an object
U = 3 nRT
2
Temperature
Suppose we heat mugs of water from 25oC to 90oC.
One Mug
25oC to 90oC
Same Temp. Change
Requires less heat
Two Mugs
25oC to 90oC
Same Temp. Change
Requires more heat
Specific Heat
Specific Heat – Amount of heat needed to raise the
temperature of one gram of a substance by one degree
Celsius or Kelvin
•
•
•
•
Unit – J/kgoC
Symbol = C
The higher the specific heat, the more energy needed to
raise the temperature
Wooden spoon versus a metal spoon
Calculating Heat
Examples:
•
•
•
•
Al has a specific heat of 0.22 kcal/kg oC
Gold has a specific heat of 0.03 kcal/kg oC
Which gets hotter sitting in the sun?
Water’s specific heat: 1.00 kcal/kg oC or
4186 J/kgoC
Calculating Heat
Suppose you immerse a hot pan into a
dishpan of water to cool it. Which will
experience a greater change in temperature
(gain or loss)?
Heat lost = -Heat gained
Q1 + Q2 + Q3 +…. = 0
Calculating Heat
Q = mCpDT
Q = heat (J)
m = mass (kg)
Cp = specific heat (J/kgoC)
DT = Tfinal – Tinitial
Calculating Heat: Example 1
How much heat must be supplied to a 500.0
gram iron pan (C = 450 J/kgoC) to raise its
temperature from 20.0oC to 100oC?
Q = mCDT
Q = (0.500 kg)(450/kg oC)(100oC -20oC)
Q = (0.500 kg)(450/kg oC)( 80oC)
Q = 18,000 J or 18 kJ
Calculating Heat: Example 2
Suppose the pan is filled with 400 g of water.
What would be the total heat?
Q = mCDT
Q = (0.400 kg)(4186/kg oC)(100oC -20oC)
Qwater =134,000 J
Qtotal = 152,000 J or 152 kJ
Calculating Heat: Example 3
200-g of tea at 95oC is poured into a 150-g glass
(C = 840 J/kgoC) at 25oC. What will be the final
temperature of the cup/glass?
Heat lost = -Heat gained
Qtea = - Qcup
mteaCteaDT = -mcupCcupDT
mteaCteaDT = -mcupCcupDT
(0.200 kg)(4186J/kgoC)(T-95oC) =
-(0.150 kg)(840J/kgoC)(T-25oC)
(837)(T-95oC) = -(126)(T-25oC)
837T – 79,500 = 3150 -126T
963 T = 82,700
T = 86oC
Calculating Heat: Example 4
A 100-g piece of aluminum (C = 900 J/kgoC) is
heated to 100 oC and immersed in 250-g of water
at 20 oC. What is the final temperature of the
system?
Heat lost = -Heat gained
Qtea = - Qcup
mAlCAlDT = -mwaterCwaterDT
mAlCAlDT = -mwaterCwaterDT
(0.100 kg)(900J/kgoC)(T-100oC) =
-(0.250 kg)(4186J/kgoC)(T-20oC)
(90)(T-100) = -(1047)(T-25)
90T – 9,000 = 20,930 - 1047T
1137T = 29,930
T = 26oC
Phase Changes
Definition - A change of state in which energy is
absorbed or released without a temperature
change.
Example: Freezing water
1. Water is cooled from 25oC to 0oC (temperature
change, not a phase change)
2. Water freezes at 0oC (no temperature change
occurs during freezing, phase change)
Phase Change
• Suppose we are heating ice:
• All of the heat goes into melting the ice rather
than increasing the temperature
• Temperature will only rise once it is all melted
Solid molecules absorb the heat to
get moving rather than increasing
temperature
Melted
molecules
now move
freely
Phase Changes: Think about it.
1. Can you get water above 100oC in a pot if you
are cooking?
2. Can you get steam above 100oC in a pressure
cooker or furnace?
3. Can you get ice below 0oC in a freezer?
Heating Curves
Boili
ng
Meltin
g
Latent Heat
No Phase Change
Q = mCpDT
Phase Change
Q = mLf
or
Q = mLv
Lf = Latent heat of fusion (freezing/melting)
Lv = Latent heat of vaporization (boiling/condensing)
Heating Curves
Temperature
(oC)
Use q = mLv
Boili
ng
Use q = mLf
Meltin
g
Use q = mCpDT
Heat (Joules)
Important Values
Here are some important values that we will need:
Substance
Steam
Water
Ice
C
2010 J/kg oC
4186 J/kg oC
2100 J/kg oC
Latent Heat of fusion (water)
Lf = 3.33 X 105 J/kg
Latent Heat of vaporization(water) Lv = 22.6 X 105 J/kg
Heating Curves: Example 1
How much energy is required to cool 100.0
grams of water at 20.0oC to make ice at –
10.0oC?
There are three steps involved here:
1. Cooling the water
2. Freezing the water
3. Cooling the ice down to –10.0oC
Let’s look at the graph:
Temperature
(oC)
Freezing
(q=mLf)
Water cools
(q=mCpDT)
Ice cools
(q=mCpDT)
Heat (Joules)
Let’s look at it step by step
1. Cooling the water
Q = mCpDT = (0.100 kg)(4186J/kgoC)(0oC20oC)
Q = 837 J
2. Freezing the water
Q = mLf = (0.100 g)(3.33 X 105 J/kg)= 3.33
X 104 J
3.
Cooling the ice down to –10.0oC
q = mCpDT = (0.100 kg)(2100J/kgoC)(-10oC-0oC)
q = 2100 J
Lastly, we just add the three heat values together:
837J + 33,000J + 2100J = 36,200 J
Heating Curves: Example 2
How much energy does a refrigerator have to
remove from 1.5 kg of water at 20 oC to make
ice at -12 oC?
ANS: 660 kJ
Calorimeter
• Foam cup
• Insulated metal container (thermos)
• Must consider heat absorbed by the calorimeter
in the equations
Calorimeter: Example 1
A 0.150-kg sample of a new alloy is heated to 540
oC. It is placed into 400-g of water at 10.0 oC
which is contained in a 200-g Aluminum
calorimeter. The final temperature of the
mixture is 30.5 oC. Calculate the specific heat of
the new alloy.
Heat lost = - Heat gained
malloyCalloyDT = -[mwaterCwaterDT + mcalCcalDT]
mwaterCwaterDT = (0.400kg)(4186 J/kg oC)(30.5 oC –
10.0oC)
mwaterCwaterDT = 34,300 J
mcalCcalDT = (0.200kg)(900 J/kg oC)(30.5 oC –
10.0oC)
mcalCcalDT = 3700 J
malloyCalloyDT = -[mwaterCwaterDT + mcalCcalDT]
malloyCalloyDT = -[34,300 J + 3700 J]
malloyCalloyDT = -38,000 J
(0.150-kg)(Calloy)(10 oC – 540 oC) = -38,000 J
(-79.5 kg oC)(Calloy) = -38,000 J
Calloy = 478 J/ kg oC
Example 2
100 g of ice at -20oC is added to 500 g of soad at
20oC.
a. Calculate the heat required to raise the
temperature of the ice to the melting point.
b. Calculate the heat required to melt the ice.
c. Is there enough heat energy available if the
soda cools to 0oC?
d. Calculate the final temperature of the soda
We will skip specific heats of gases
Heat Flow
1. Conduction
2. Convection
3. Radiation
Conduction
Conduction – The transfer of energy from molecule to
molecule through collisions.
1. Warmer (faster) molecules to slower(colder)
molecules
2. Thermometers work through conduction.
3. Depends on the area of contact.
Conduction: thermometer
Slower
(colder)
mercury
atoms get
“bumped
” and
accelerat
ed by the
collisions
with the
air
The warmer, faster moving air
molecules collide with the glass and
give some of their kinetic energy of
motion to the glass and mercury.
Conduction
4. Air is a poor conductor - molecules are so far apart,
far fewer collisions than in a solid.
5. Insulators usually have a number of air pockets
(like in home insulation) to slow the heat flow.
6. Cooking on top of a stove is usually conduction
since the pan/pot is in contact with a hotter burner
element or a flame.
Conduction
DQ = kA(T1 – T2)
Dt
L
DQ
Dt
k
A
T1 – T2
L
= Rate of heat flow (heat/time)
= Thermal Conductivity
= Area
= Change in Temperature
= Length
Conduction: Example 1
Calculate the rate of heat flow through a window 3
m2 and 3.2 mm thick. The outer and inner
temperatures are 14.0 oC and 15.0 oC
Conduction: R-values
• R values – commercial measure of the quality of
insulation
• Higher R = more insulation
• R = length
k
R
Glass
1
Brick
1
Insulation
12-20
Convection
Convection – the transfer of hear by the
movement of a large volume of air or liquid
1. Hot-air home heating
2. Radiator wars air that it comes in contact with
through conduction. Convection takes over as it
moves through the room
3. Convection ovens
Convection
Then convection causes the warm
air to expand into the room (the
hot air rises)
The cold air sinks and comes in
contact with the radiator again.
A hot
radiator
warms air
through
conductio
n (contact
between
the atoms/
molecules
Radiation
Radiation – The transfer of heat through
electromagnetic waves
1. Light (microwaves, infrared, visible, etc..)
2. The waves give kinetic energy to molecules and
cause them to move faster (increase in
temperature).
3. Sun
4. Red heat lamps at fast-food places.
Radiation
Light and other radiation from the sun
strike the earth impart some of their
energy to the molecules on the earth
Earth
What type of heat transfer is occurring
in each picture?
http://en.wikipedia.org/wiki/Image:Fireplace_Burning.jpg
Radiation
Stefan-Boltzmann Equation
DQ = esA(T14 – T24)
Dt
s = 5.67 X 10-8 W/m2 K4 (Stefan-Boltzmann
constant)
A = area
T = Temperature (kelvin)
e = Emissivity
Radiation: Emissivity
e = Emissivity
• Value between 0 and 1
• 1
– Black substances
– Absorb and emit radiation well
• 0
– Shiny substances
– Do not absorb or emit well
Radiation: Example 1
An athlete is sitting in a locker room at is 15 oC.
The athlete has a skin temperature of 34 oC and
an e value of 0.7. If his surface area is 1.5 m2,
calculate the rate of heat loss.
DQ = esA(T14 – T24)
Dt
DQ = (0.7)(5.67 X 10-8 W/m2 K4)(1.5m2)(3074 – 2884)
Dt
DQ/Dt = 120 W
Radiation: Example 2
A ceramic teapot (e=0.70) and a shiny teapot (e =
0.10) filled with tea at 95 oC. Calculate the rate
of heat loss of each if the room is 20oC. Assume
each teapot is about 0.05 m2.
ANS: 22 W, 3.1 W
Radiation and Sun Angle
• Get less radiation as the sun becomes less
vertical
• Would you absorb more sun at midday or 6 PM?
• Would you absorb more sun in Mexico or Ohio?
Solar constant = 1350 W/m2 (about 1000 when
adjusted for the air)
DQ = (1000 W/m2)eA cos q
Dt
Sun: Example 1
What is the energy absorption of a person (0.80 m2 , e =
0.70) getting a tan if the sun makes an angle of 30o with
the vertical.
DQ = (1000 W/m2)eA cos q
Dt
DQ = (1000 W/m2)(0.7)(0.80m2) cos 30o
Dt
DQ/Dt = 490 W (wearing light clothes will reduce e)
Thermal Pollution
• Thermally polluting power plants
– Coal plants
– Oil plants
– Nuclear plants
• Non-thermally polluting power plants
–
–
–
–
Hydroelectric
Tidal energy
Wind
Solar
Work: Example 4
Determine the change in internal energy of 1.0 liter
of ware at 100oC when it is fully boiled. It
results in a 1671 L of steam at 100oC. Assume
the process is done at atmospheric pressure.
DU = Q – W
Q = mL =(1.0 kg)(22.6 X 105 J/kg) = 22.6 X 105 J
W = PDV
(1 L = 1 X 10-3 m3)
W = (1 X 10 5 N/m2)(1671 X 10-3 m3 - 1X 10-3 m3)
W = 1.7 X 105 J
DU = Q – W
DU = 22.6 X 105 J – 1.7 X 105 J
DU = 21 X 105 J (or 2.1 X 106 J)
Work: Example 2
Determine the change in internal energy of 1.0 liter
of ware at 100oC when it is fully boiled. It
results in a 1671 L of steam at 100oC. Assume
the process is done at atmospheric pressure.
DU = Q – W
Q = mL =(1.0 kg)(22.6 X 105 J/kg) = 22.6 X 105 J
W = PDV
(1 L = 1 X 10-3 m3)
W = (1 X 10 5 N/m2)(1671 X 10-3 m3 - 1X 10-3 m3)
W = 1.7 X 105 J
DU = Q – W
DU = 22.6 X 105 J – 1.7 X 105 J
DU = 21 X 105 J (or 2.1 X 106 J)