1. Zeroth and First Law
of Thermodynamics
Introduction to Thermodynamics

2. Learning Outcomes
 You will
 State and explain the Zeroth Law of Thermodynamics
 State the First Law of Thermodynamics and derive mathematical
relationship for a closed system
 Derive relationships for Internal Energy Change and Enthalpy
Change in terms of specific heat capacitates of the substance and
temperature change
 Derive relationship between specific heat capacities, and the
Universal gas constant for ideal gasses
 Use First Law of thermodynamics together with ideal gas equation
to solve simple problems

3. Content
 Zeroth Law of Thermodynamics
 First Law of Thermodynamics
 Derivations from First Law
 Internal Energy and specific heat capacity under constant volume
 Enthalpy and specific heat capacity under constant pressure
 Relationship between specific heat capacities for ideal gasses
 Examples

4. Zeroth Law of Thermodynamics
 Let A, B and C (say is a thermometer) be three systems. If A and B are
separately in thermal equilibrium with C, then A and B are in thermal
equilibrium.
Thermal equilibrium: Two systems are in thermal equilibrium if they
could transfer heat between each other, but don't.
In Thermal
equilibrium

5. Zeroth Law of Thermodynamics
 As a consequence of Zeroth law we can conclude that
“Temperature” a property that determines whether an
object is in thermal equilibrium with other objects
 Two objects in thermal equilibrium with each other are at
the same temperature
 If two objects have different temperatures, they are not in
thermal equilibrium with each other

6. First Law of Thermodynamics
 First law of thermodynamics states that energy can neither
be created nor destroyed, it can only change the form.
 For a closed system or a fixed mass:
Net energy
transferred to the
system as heat and
work
Net increase in
the total energy
of the system=

7. WQKEPEUKEPEU
WQEE


)()( 111222
12
WQKEKEPEPEUU  )()()( 121212
KEPEUWQ 
UWQ  For a stationary closed systems
Sign convention
Work done by the system - Positive
Heat supplied to the system - Positive
First Law of Thermodynamics

8. Example 1
A certain amount of air is compressed in a cylinder. The
change in internal energy of the air is 10kJ. Work required for
the compression is 500kJ. What is the amount of heat
transfer?
UWQ 
Work done by the system - Positive
Heat supplied to the system - Positive
kJWUQ 45050010 
Sign convention
Heat is rejected
Solution

9. Derivations from First Law
Consider constant pressure process
UWQ 
 
HQ
pVUpVUQ
VVpUUQ
VpUUQ




)()(
)(
1122
1212
12
HTCQ p 
Specific heats at constant volume and constant pressure
If Cp is Specific Heat Capacity of the substance;
TmcQ P

10. Derivations from First Law
Consider constant volume process
UWQ 
Since there is no work transfer, UQ 
If Cv is Specific Heat Capacity of the substance under constant volume ;
TmcQ v
Therefore; UTCQ v 

11. Derivations from First Law
pVUH 
mRTUH 
TmRUH 
TmRTCTC vp 
mRCC vp 
Rcc vp 

v
p
c
c
= specific heat ratio
This is equal to 1.4 many adiabatic
process
Specific heat relations for ideal gases

12. Example 2
A 0.6kg copper piece at 100oC is dropped in to an insulated tank
containing 0.75kg of liquid water at 25oC.
Determine the temperature of the system when thermal
equilibrium is reached. Specific heats of copper and water
are:0.393kJ/kgK and 4.184kJ/kgK
For incompressible substances (solids and liquids) cp and cv
take same value

13. Take the copper piece and water as the system. The temperature
after equilibrium is reached is T2.
UWQ 
U0
0 wcop UU
0)()( 1212  wwwcopcopcop TTcmTTcm
0)25)(184.4(75.0)100)(393.0(6.0 22  TT
CT o
24.302 
Solution

14. Example 3
The temperature of 3.5kg of gas in a container is increased from
22oC to 39oC by heating it. The heat transferred during the process
is 25kJ. Calculate, treating the gas to be a perfect gas,
(a) change in internal energy and
(b) work done
The specific heats ratio and the molar mass of the gas are 1.4 and
28g respectively.

15. UWQ  kJQ 25
 12
1
TT
mR
TCU v 



1
,1
1






mR
C
C
mR
C
mR
C
C
mRCC
v
v
vv
p
vp
 
)2239(
14.1
)028.0/314.8)(5.3(
1
)/(
12





 TT
MRm
U o

kJU 17.44
17.4425 W
kJW 17.19 Minus means work has to be done on the system.
This implies that in order to increase the internal energy by 44.17kJ, 25kJ is not sufficient
Solution

16. Example 4
An insulated cylinder closed at both ends contains free moving
piston. One side of the cylinder is filled with nitrogen, and the other
side with air. Initial pressure and volume of each gas are 1bar and
0.01m3 respectively. The initial temperature is 20oC for both the
gasses.
The air in the cylinder is heated by means of an electrical heater
installed inside the cylinder until the volume occupied by the
nitrogen is reduced to 0.008m3
Find
(a) final temperature of air and
(b) Work done by air
(c) heat added to air
Cv for air is 0.718kJ/kgK and R for air is 0.287kJ/kgK. Adiabatic index for
nitrogen is 1.4

17. Applying ideal gas equation for air state 2 )1.......(222 aaaaa TRmVp 
Applying ideal gas equation for air state 1 )2........(111 aaaaa TRmVp 
From (1)/(2)
aa
aa
a
a
Vp
Vp
T
T
11
22
1
2
 )3........(1
11
22
2 a
aa
aa
a T
Vp
Vp
T 
3
111 01.0,29327320,1 mVKTbarp aaa 
(a) Final temperature of air
Solution

18. Now find V2a NaNa VVVV 2211 
008.001.001.0 2  aV
3
2 012.0 mV a 
Now find P2a
Under equilibrium condition , P2a=P2N

NNNN VpVp 2211  Because nitrogen gas undergoes
adiabatic process for which
Polytropic index is 1.4
barbar
V
Vp
p
N
NN
N 367.1
008.0
01.0
)1(
4.1
2
11
2 





 


19. KT a 6.480)293(
)01.0)(1(
)012.0)(367.1(
2 
From equation (3) )3........(1
11
22
2 a
aa
aa
a T
Vp
Vp
T 
(b) Work done by air
4.11
)01.0)(1()008.0)(367.1(
1
1122







NNNN
N
VpVp
W
kJWN 234.0
kJWa 234.0

20. kJkJWUQ 254.12234.002.16 
UWQ 
 aava
aa
aa
aaav TTc
TR
Vp
TTmcU 12
1
11
12 )( 
  kJU 02.162936.480)718.0(
)293)(287.0(
)01.0)(10)(01.0( 5

Using first law
(C) Heat added to air

21. Example 5
In a piston cylinder device, 300g of saturated water vapor
maintained at 200kPa is heated by a resistance heater installed
within the cylinder for 10min by passing current of 0.35A from
220V source. The heat loss from the system during the heating
process is 2.2kJ.
Calculate (a) final temperature of the steam
(b) work done

22. )1.....(UWQ 
kJVItQE 2.46)600)(35.0)(220( 
kJQ 442.22.46 
)2)....(( 12 hhmHQ 
kgkJh /5.27061 
From steam tables, at 200kPa and sat.
vapour. condition
(a) Final temperature

23. )5.2706(3.044),2( 2  hFrom
kgkJh /16.28532 
From steam tables, when h2 = 2853.16 at 200kPa
CT o
47.1912 
At 200kPa
T h
150 2768.8
T2 h2=2853.16
200 2870.5
8.27685.2870
8.276816.2853
150200
1502




T

24. )( 12 uumU 
kgkJu /5.25291 
From steam tables, at 200kPa and sat.
vapour. condition
U2 is the internal energy at 200kPa and at
191.47oC
At 200kPa
T u
150 2576.9
191.4
7
u2
200 2654.4
9.25764.2654
9.2576
150200
15047.191 2




 u
kgkJu /2.26412 
)5.25292.2641(3.0 U
kJU 51.33
From (1)
UWQ 
kJW
W
49.10
51.3344


(b) Work done

25. END