The First Law of
Thermodynamics
Closed System
1
So far…

We’ve considered various forms of energy

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Q (heat)
W (work)
E (Energy)
We haven’t tried to relate them during a
process
They are related through the
FIRST LOW OF THERMODYNAMICS
(CONSEVATION OF ENERGY PRINCIPLE)
2
Recall…

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Energy can be neither created nor destroyed;
it can only change forms ( a rock falling off
the cliff)
Consider a system undergoing a series of
adiabatic processes from a specified state 1
to another specified state 2.
Obviously the processes can not involve any
heat transfer but they may involve several
kinds of work interactions
3
Recall…
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Experimentally observed: all adiabatic
processes between two specified states of a
closed system, the net work done is the same
regardless of the nature of the closed system
and the details of the process.
This principle is called the first law
A major consequences of the first law is the
existence and the definition of the property
total energy E
4
Recall…
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The net work is the same for all adiabatic processes
of a closed system between two specified states,
The net work must depend on the end states of the
system only.
Therefore it must correspond to a change in
property of the system (i.e. the total energy)
Simply the change in the total energy during an
adiabatic process must be equal to the net work
done.
5
First low of thermodynamics
for Closed Systems

Reminder of a Closed System.
 Closed
system = Control mass
 It is defined as a quantity of
matter chosen for study.
 No mass can cross its boundary
but energy can.
6
Experience 1: W= 0
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If we transfer 5 kJ of heat to a potato,
its total energy will increase by 5 kJ.
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Experience 2: W=0

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Heat a water in a pan by transferring 15 KJ from
the range
There is 3 KJ losses
This means E2-E1= 12 KJ
8
Experience 3: Q=0

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Heat an insulated room with electric heater.,
W= 5KJ (electric)
means E2-E1= 5KJ.
Replace the electric
heater
with a paddle wheel.
9
Experience 4: Q=0

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If we do a 10 kJ of boundary
work on a system,
the system’s internal energy
will increase by 10 kJ.
This is because (in the
absence of any heat transfer (Q
= 0), the entire boundary work
will be stored in the air as part
of its total energy.
10
Experience 5: Qin, Qout, Win

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If we do 6 kJ of shaft
work on system and
Transfer 15 kJ of heat
in.
Loose 3 kJ of heat
out.
Doing 4 kJ of Work
out.
Then the system
internal energy will
increase by 14 kJ.
Wb= 4 kJ
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Before proceeding further, let us fix
the direction of Energy Transfer
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Heat transferred in to the system is positive
 Qin is +ve
Heat transferred out of the system is negative
 Qout is -ve
Work done on the system increases energy of the
system.
 Win is +ve
Work done by the system decreases energy of the
system.
 Wout is -ve
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Based on the previous experimental
observations, the conservation of energy
principle may be expressed as:
 Total Energy
  Total Energy
  Changein the total 

  
  

 Entering the System   Leaving the System   system energy

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Ein  Eout  Esys
On a rate basis
E in  E out  E sys
Rate of net energy transfer
by heat, work and mass
Rate of change in total
energy of the system
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Let us discuss the right hand side of the
first low equation, i. e.  E
E  U  KE  PE
Usually the KE and PE are small
E  U  mu
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Let us turn to the Left hand side of the 1st
low equation
Ein  Eout  E
Eout
Ein
0
Qin  Win  Emass,in  ( Qout  Wout  Emass,out )  E
0
Emass=0 for closed system
Qin  Win  ( Qout  Wout )  E
If we rearrange, we get
Qin  Qout  Win  Wout  E
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The First low of thermodynamics
Qin  Qout  Win  Wout  E
Assuming Qin>Qout and Wout> Win
Qnet ,in  Wnet ,out  E
usually we drop the subscripts, hence
Q  W  E
Q  W  E
General Form (KJ)
per unit time (or on a Rate basis) KJ/s= Watt
q  w  e
per unit mass basis (KJ/kg)
q  w  de
differential form
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KEY CONCEPT
Q  W  E
17
Energy Change for a system
undergoing a cycle

Q-W =U

But U = U2-U1=0
(initial state and final
states are the same)

Hence : Q=W
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Example (4-1):
Cooling of a Hot fluid in a Tank
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A rigid tank contains a hot fluid that is cooled
while being stirred by a paddle wheel. Initially,
the internal energy of the fluid is 800kj. During
the cooling process, the fluid loses 500kj of
heat, and the paddle wheel does 100 kj of work
on the fluid. Determine the final internal energy
of the fluid.
Assumptions:
Tank is stationary and thus KE=PE=0.
Analysis:
Q-W=U2-U1
19
Example (4-2):
Electric Heating of a Gas at Constant Pressure
• A piston-cylinder device contains 25 g of
saturated water vapor that is maintained at
a constant pressure of 300 kPa. A
resistance heater within the cylinder is
turned on and passes a current of 0.2 A for
5 min from a 120-V source. At the same
time, a heat loss of 3.7 kJ occurs.
• Show that for a closed system the
boundary work Wb and the change in
internal energy U in the first-law
relation can be combined into one
term, H, for a constant pressure
process.
• Determine the final temperature of the
steam.
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Example (4-3):
Unrestrained Expansion
of water into an
Evacuated Tank
A rigid tank is divided into two equal parts
by a partition. Initially, one side of the tank
contains 5 kg of water at 200 kPa and 25oC,
and the other side is evacuated. The
partition is then removed, and the water
expands into the entire tank. The water is
allowed to exchange heat with its
surroundings until the temperature in the
tank returns to the initial value of 25oC.
Determine a) the volume of the tank, b) the
final pressure, and c) the heat transfer for
this process.
Expansion a against a vacuum
involves no work and thus no
energy transfer
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Example (4-6):
Heating of a
Gas at
Constant
Pressure
A piston-cylinder device initially contains air at 150 kPa
and 27oC. At this state, the piston is resting on a pair of
stops, as shown in the figure above, and the enclosed
volume is 400 L. The mass of the piston is such that a
350-kPa pressure is required to move it. The air now
heated until its volume has doubled. Determine (a) the
final temperature, (b) the work done by the air, and c)
the total heat transferred to the air. <Answers: a) 1400 K, b)
140 kJ, c) 766.9 kJ>
Sol:
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