Thermodynamics principles for engineering systems

F1015B: Aplicación de la
termodinámica en sistemas
ingenieriles
M2. Calor y leyes de la termodinámica

There is two concepts that we should know:
1. Thermal contact
2. Thermal equilibrium

Two objects are said to be in thermal contact if it can exchange energy
through the process of heat
• Conducción
• Convección
• Radiación

Temperature and the Zeroth Law of Thermodynamics
If objects A and B are separately in thermal equilibrium with a third
object C, then A and B are in thermal equilibrium with each other.

Two objects, with different sizes, masses, and temperatures, are
placed in thermal contact. In which direction does the energy travel?
(a) Energy travels from the larger object to the smaller object.
(b) Energy travels from the object with more mass to the one with
less mass.
(c) Energy travels from the object at higher temperature to the object
at lower temperature.

Two objects, with different sizes, masses, and temperatures, are
placed in thermal contact. In which direction does the energy travel?
(a) Energy travels from the larger object to the smaller object.
(b) Energy travels from the object with more mass to the one with
less mass.
(c) Energy travels from the object at higher temperature to the
object at lower temperature.

Thermometers and the Celsius Temperature Scale

Thermometers
Thermometers are based on principle that some physical
property of system changes as system’s temperature changes:
• Volume of liquid
• Dimensions of solid
• Pressure of gas at constant volume
• Volume of gas at constant pressure
• Electric resistance of conductor
• Color of object

The Constant-Volume Gas Thermometer
0
P P gh

= +

The Constant-Volume Gas Thermometer
and the Absolute Temperature Scale
C 273.15
T T
= −

The Absolute Temperature Scale

The Celsius, Fahrenheit, and Kelvin Temperature Scales
F C
9
32 F
5
T T
= + 
C F
5
9
T T T
 =  = 
C 273.15
T T
= −

F C
9
32 F
5
T T
= + 
C F
5
9
T T T
 =  =  C 273.15
T T
= −
For equations that need temperature T or involves ratio of temperatures:
• Convert all temperatures to kelvins
• For equations containing change in temperature ΔT
• Using Celsius temperatures will give you correct answer
Always safest to convert temperatures to Kelvin scale

Consider the following pairs of materials. Which pair represents two
materials, one of which is twice as hot as the other?
(a) boiling water at 100C, a glass of water at 50C
(b) boiling water at 100C, frozen methane at −50C
(c) an ice cube at −20C, flames from a circus fire-eater at 233C
(d) none of those pairs

Thermal Expansion of Solids and Liquids

Thermal Expansion of Solids
/ i
L L
T




i
L L T

 = 
( )
f i i f i
L L L T T

− = −

Coefficients of Linear Expansion

Volume Expansion
i
V V T

 = 
( )( )( )
( )( )( )
( )
( ) ( )
3
2 3
1
1 3 3
i
i
V V w w h h
T w w T h h T
wh T
V T T T
  

  
+  = +  +  + 
= +  +  + 
= + 
 
= +  +  + 
 
( ) ( )
2 3
/ 3 3
i
V V T T T
  
 =  +  + 
( )
/ 3 3
i i
V V T V V T
 
 =    = 
3
 
=

29
Example
A segment of steel railroad track has a length of 30.000 m when the
temperature is 0.0C.
(A) What is its length when the temperature is 40.0C?
30

30
Example
A segment of steel railroad track has a length of 30.000 m when the
temperature is 0.0C.
(A) What is its length when the temperature is 40.0C?
( ) ( )( )
1
6
11 10 C 30.000 m 40.0 C 0.013 m
i
L L T

−
−
 = 
 
=    =
 
30.000 m 0.013 m 30.013 m
f
L = + =

31
Example
What if the temperature drops to −40.0C? What is the length of the
unclamped segment?

32
Example
What if the temperature drops to −40.0C? What is the length of the
unclamped segment?
𝐿𝑓 = 30.000 m − 0.013 m = 29.987 m

Macroscopic Description of an Ideal Gas

Ideal Gas Law
m
n
M
=
PV nRT
=
8.314 J/mol K
R = 
0.08206 L atm/mol K
R =  

Boltzmann’s Constant
B
A
N
PV nRT RT
N
PV Nk T
= =
=
23
B 1.38 10 J/K
A
R
k
N
−
= = 
PV nRT
=

38
Example
A spray can containing a propellant gas at twice atmospheric
pressure (202 kPa) and having a volume of 125.00 cm3 is at 22C. It
is then tossed into an open fire. (Warning: Do not do this
experiment; it is very dangerous.) When the temperature of the gas
in the can reaches 195C, what is the pressure inside the can?
Assume any change in the volume of the can is negligible.

39
Example
A spray can containing a propellant gas at twice atmospheric
pressure (202 kPa) and having a volume of 125.00 cm3 is at 22C. It
is then tossed into an open fire. (Warning: Do not do this
experiment; it is very dangerous.) When the temperature of the gas
in the can reaches 195C, what is the pressure inside the can?
Assume any change in the volume of the can is negligible.
PV
PV nRT nR
T
=  = f f
i i
i f
P V
PV
T T
=
f
i
i f
P
P
T T
 =
( )
468 K
202 kPa 320 kPa
295 K
f
f i
i
T
P P
T
   
= = =
   
 
 

Heat and Internal Energy
Internal energy is all the energy of a system that is
associated with its microscopic components—atoms
and molecules—when viewed from a reference frame at
rest with respect to the center of mass of the system.
Heat is defined as a process of transferring energy
across the boundary of a system because of a
temperature difference between the system and its
surroundings. It is also the amount of energy Q
transferred by this process.

Units of Heat
calorie (cal): energy transfer necessary to raise
temperature of 1 g of water from 14.5C to 15.5C
British thermal unit (Btu): energy transfer required to
raise temperature of 1 lb of water from 63°F to 64°F
Calorie (Cal): equivalent to 1 000 calories

The Mechanical Equivalent of Heat
int MW
E W T
 = +

Joule’s Experiment
1 cal 4.186 J
=

45
Example
Losing Weight the Hard Way
A student eats a dinner rated at 2 000 Calories. He wishes to do an
equivalent amount of work in the gymnasium by lifting a 50.0-kg
barbell. How many times must he raise the barbell to expend this
much energy? Assume he raises the barbell 2.00 m each time he lifts
it and he transfers no energy when he lowers the barbell.

46
Example
total total
U W
 = U mgh
 =
total
U nmgh
 =
total
nmgh W
=
total
W
n
mgh
=
( )( )( )
3
2
3
2000 Cal 1.00 10 cal 4.186 J
Calorie 1 cal
50.0 kg 9.80 m/s 2.00 m
8.54 10 times
n
 
  
=   
 
 
= 

Specific Heat and Calorimetry
Q C T
= 

Specific Heat
Q
c
m T


Q mc T
= 
( )( )( ) 3
0.500 kg 4186 J/kg C 3.00 C 6.28 10 J
Q =   = 
int ER
Examples
potato in microwave:
E T mc T
 = = 
int
bicycle pump:
E W mc T
 = = 

Specific Heat
Imagine you have 1 kg each of iron, glass, and water, and all three samples
are at 10°C.
Rank the samples from highest to lowest temperature after 100 J of energy is
added to each sample.

Specific Heat
are at 10°C.
Rank the samples from highest to lowest temperature after 100 J of energy is
added to each sample.
iron, glass, water

Specific Heat
are at 10°C.
Rank the samples from greatest to least amount of energy transferred by heat
if each sample increases in temperature by 20°C.

Specific Heat
are at 10°C.
Rank the samples from greatest to least amount of energy transferred by heat
if each sample increases in temperature by 20°C.
water, glass, iron

Calorimetry
cold hot
Q Q
= −
( )
( )
w w f w
x x f x
m c T T
m c T T
−
= − −
( )
w w f w
m c T T
−
( )
x x f x
m c T T
− −

56
Example
A 0.050 0-kg ingot of metal is heated
to 200.0°C and then dropped into a
calorimeter containing 0.400 kg of
water initially at 20.0°C. The final
equilibrium temperature of the mixed
system is 22.4°C. Find the specific heat of the metal.

57
Example
A 0.050 0-kg ingot of metal is heated
to 200.0°C and then dropped into a
calorimeter containing 0.400 kg of
water initially at 20.0°C. The final
equilibrium temperature of the mixed
system is 22.4°C. Find the specific heat of the metal.
( )
( )
w w f w
x
x x f
m c T T
c
m T T
−
=
−
( )( )( )
( )( )
0.400 kg 4186 J/kg C 22.4 C 20.0 C
453 J/kg C
0.050 0 kg 200.0 C 22.4 C
x
c
  − 
= = 
 − 

58
Example
453 J/kg C
= 
x
c

Latent Heat
Q
L
m


Q L m
= 

Latent Heat
( )( )( )
3
ice ice 1.00 10 kg 2090 J/kg C 30.0 C 62.7 J
−
=  =    =
Q m c T

Latent Heat
( )( )
5 3
ice 3.33 10 J/kg 1.00 10 kg 333 J
−
=  = =   =
f w f
Q L m L m

Latent Heat
( )( )( )
3 3
1.00 10 kg 4.19 10 J/kg C 100.0 C 419 J
w w
Q m c T −
=  =     =

Latent Heat
( )( )
6 3 3
2.26 10 J/kg 1.00 10 kg 2.26 10 J
v s v w
Q L m L m −
=  = =   = 

Latent Heat
( )( )( )
3 3
1.00 10 kg 2.01 10 J/kg C 20.0 C 40.2 J
s s
Q m c T −
=  =     =

Latent Heat
3 3
total 62.7 J 333 J 419 J 2.26 10 J 40.2 J 3.11 10 J
Q = + + +  + = 

68
Example
What mass of steam initially at 130°C is needed to warm 200 g of water in a 100-g glass container from 20.0°C to
50.0°C?

69
Example
50.0°C?
cold hot
Q Q
= −
1 s s s
Q m c T
= 
( )
2 0
v s v s s v
Q L m L m m L
=  = − = −
3 hot water
s w
Q m c T
= 
cold cold water glass
w w g g
Q m c T m c T
=  + 
( )
hot 1 2 3 hot water
s s s v w
Q Q Q Q m c T L c T
= + + =  − + 

70
Example
50.0°C?
( )
cold water glass hot water
w w g g s s s v w
m c T m c T m c T L c T
 +  = −  − + 
cold water glass
hot water
w w g g
s
s s v w
m c T m c T
m
c T L c T
 + 
= −
 − + 
cold cold water glass
w w g g
Q m c T m c T
=  + 
( )( )( ) ( )( )( )
( )( ) ( ) ( )( )
6
2
0.200 kg 4186 J/kg C 50 C 20.0 C 0.100 kg 837 J/kg C 50.0 C 20.0 C
210 J/kg C 100 C 30 C 2.26 10 J/kg 4 186 J/kg C 50.0 C 20.0 C
1.09 10 kg 10.9 g
s
m
−
  −  +   − 
= −
  −  −  +   − 
=  =

Thermal Conduction

= 

T
Q kA t
x
dT
P kA
dx
=

Thermal Conduction
h c
T T
dT
dx L
−
=
h c
T T
P kA
L
−
 
=  
 
( )
( )
/
h c
i i
i
A T T
P
L k
−
=


75
Example
Determine the temperature at the
interface and the rate of energy
transfer by conduction through an area A of the slabs in
the steady-state condition.
Two slabs of thickness L1 and L2
and thermal conductivities k1 and
k2 are in thermal contact with each
other as shown in the figure. The
temperatures of their outer surfaces
are Tc and Th, respectively, and
Th > Tc.

76
Example
1 1
1
c
T T
P k A
L
 
−
=  
 
2 2
2
h
T T
P k A
L
 
−
=  
 
1 2
1 2
c h
T T T T
k A k A
L L
   
− −
=
   
   
1 2 2 1
1 2 2 1
c h
k L T k L T
T
k L k L
+
=
+
( )
( ) ( )
1 1 2 2
/ /
h c
A T T
P
L k L k
−
=
+

Home Insulation
( )
h c
i
i
A T T
P
R
−
=

( )
( )
/
h c
i i
i
A T T
P
L k
−
=

R-value of a typical 1 in-thick Vacuum insulated panel
(VIP) would be 20 h·ft2·°F/BTU.

78
Example
Calculate the total R-value for a
wall constructed as shown in the
figure. Starting outside the house
(toward the front in the figure)
and moving inward, the wall
consists of 4 in. of brick, 0.5 in.
of sheathing, an air space 3.5 in.
thick, and 0.5 in. of drywall.

79
Example
( )
( )
( )
( )
( )
( )
2
1
2
2
2
3
2
4
2
5
6
outside stagnant air layer 0.17 ft F h/Btu
brick 4.00 ft F h/Btu
sheathing 1.32 ft F h/Btu
air space 1.01 ft F h/Btu
drywall 0.45 ft F h/Btu
inside stagnant air layer
0.17
R
R
R
R
R
R
=  
=  
=  
=  
=  
= 2
ft F h/Btu
 
total 1 2 3 4 5 6
2
7.12 ft F h/Btu
R R R R R R R
= + + + + +
=  

Radiation
https://www.youtube.com/watch?v=-CIUZBBz0Uc

Radiation
( )
4 4
net 0
P Ae T T

= −

Home Insulation
R-value of a typical 1 in-thick
Vacuum insulated panel (VIP)
would be 20 h·ft2·°F/BTU.

The First Law of Thermodynamics
int
E Q W
 = +
The change in the internal energy of a system is
equal to the sum of the heat gained or lost by the
system and the work done by or on the system.

The First Law of Thermodynamics
Energy can neither be created nor destroyed in a closed system.
The total energy in a system remains constant, although it may be
converted from one form to another.

Work W Done on a Gas
ˆ ˆ
dW d F dy F dy PAdy
=  = −  = − = −
F r j j
dW PdV
= −
f
i
V
V
W PdV
= −

Work and PV Diagrams
The work done on a gas in a
quasi-static process that
takes the gas from an initial
state to a final state is the
negative of the area under
the curve on a PV diagram,
evaluated between the
initial and final states.
f
i
V
V
W PdV
= −

Work and PV Diagrams
( )
i f i
W P V V
= − − ( )
f f i
W P V V
= − −
f
i
V
V
W PdV
= −

Cyclic Processes
( )
int 0 and
cyclic process
E Q W
 = = −

Isobaric Processes
( )
( )
isobaric process
f i
w P V V
= − −

Isovolumetric Processes
( )
int
isovolumetric process
E Q
 =

Isothermal Processes
constant
PV nRT
= =
int 0
E
 =
Q W
= −

Isothermal Processes
f
i
V
V
nRT
dV
V
= −
( )
ln
isothermal process
i
f
V
W nRT
V
 
=  
 
 
f
i
V
V
W P dV
= −
ln
f f
i
i
V V
V
V
dV
W nRT nRT V
V
= − = −


Adiabatic Processes
( )
int
adiabatic process
E W
 =
constant
PV 
=

102
Example
A 1.0-mol sample of an ideal gas
is kept at 0.0°C during an
expansion from 3.0 L to 10.0 L.
(A) How much work is
done on the gas during
the expansion?

103
Example
A 1.0-mol sample of an ideal gas
is kept at 0.0°C during an
expansion from 3.0 L to 10.0 L.
(A) How much work is
done on the gas during
the expansion?
( )( )( ) 3
ln
3.0 L
1.0 mol 8.31 J/mol K 273 K ln 2.7 10 J
10.0 L
i
f
V
W nRT
V
 
=  
 
 
 
=  = − 
 
 

104
Example
A sample of an ideal gas goes through the process shown in Figure. From A
to B, the process is adiabatic; from B to C, it is isobaric with 100 kJ of energy
entering the system by heat. From C to D, the process is isothermal; from D
to A, it is isobaric with 150 kJ of energy leaving the system by heat.
Determine the difference in internal energy EintB - EintA.

105
Example
A sample of an ideal gas goes through the process shown in Figure. From A
to B, the process is adiabatic; from B to C, it is isobaric with 100 kJ of energy
entering the system by heat. From C to D, the process is isothermal; from D
to A, it is isobaric with 150 kJ of energy leaving the system by heat.
Determine the difference in internal energy EintB - EintA.

106
Example
Air at 20.0C in the cylinder of a diesel engine is compressed from an initial
pressure of 1.00 atm and volume of 800.0 cm3 to a volume of 60.0 cm3.
Assume air behaves as an ideal gas with  = 1.40 and the compression is
adiabatic. Find the final pressure and temperature of the air.

107
Example
Air at 20.0C in the cylinder of a diesel engine is compressed from an initial
pressure of 1.00 atm and volume of 800.0 cm3 to a volume of 60.0 cm3.
Assume air behaves as an ideal gas with  = 1.40 and the compression is
adiabatic. Find the final pressure and temperature of the air.
( )
1.40
3
3
800.0 cm
1.00 atm 37.6 atm
60.0 cm
i
f i
f
V
P P
V

   
= = =
   
   
 
( )( )
( )( )
( )
3
3
37.6 atm 60.0 cm
293 K
1.00 atm 800.0 cm
826 K 553 C
f f
f i
i i
P V
T T
PV
= =
= = 

Thermodynamics principles for engineering systems

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