Download Lecture 1: Energy

Lecture 1: Energy
• Reading: Zumdahl 9.1
• Outline
– Energy: Kinetic and Potential
– System vs. Surroundings
– Heat, Work, and Energy
• Problems:
– Z9.10, Z9.15, Z9.18, Z9.19,Z9.20, Z9.21
– Working these problems are the key to knowing
the material and testing whether you understand
the ideas.
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Energy: Kinetic vs. Potential
• Potential Energy (PE)
– Energy due to position or
composition.
– Equals (mgh) in example.
m
• Kinetic Energy (KE)
h
v
– Energy due to motion.
– Equals mv2/2 in example.
• Total Energy is Always
Conserved
2
Energy: E = KE + PE
• Energy is the sum of kinetic energy and potential
energy.
1 mv 2
KE
=
– KE (Kinetic Energy)
2
– PE (Potential Energy) PE = mgh
• Energy is readily converted between these two
forms.
• If the system of interest is isolated (no exchange
with surroundings), then total energy is constant, or
conserved.
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Problem: Z9.15
Ball A (2kg) swings down (falls) a height of 10 m, and hits B (4kg), transfers all
energy to ball B which climbs a 3 m hill. What is the velocity of B at the top (of
the 3 m hill)?
How do we solve the:
1) What is the energy of A when it hits B? Hint: Potential energy is converted to
kinetic energy.
ΔE = 0 ⇒ KE = − PE = − m gz = 2 ⋅ 9.8 ⋅ 10 J ∼ 200 J
A
A
2) What is the initial Kinetic energy of B? Hint: All Energy is transferred to KE.
3) How much Potential Energy is gained in climbing 3m?
PE = mB gzB = 4 ⋅ 9.8 ⋅ 3J
∼ 120 J
4) What is the total energy of B? Hint: Energy is conserved.
5) What is the KE of B at the top of the hill?
E = KE + PE
200 = KE + 120 J
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Ex. for Conserved Energy: Mass on a Spring
Units of Energy in SI system
is the Joule, J. 1J ≡ kg − m 2
sec 2
• Initial PE = 1/2 kx2
• At x = 0:
0
E = KE + PE
– PE = 0
– KE = 1/2mv2
= mv + kx
1
2
• Example:
2
– Init. PE = 10 J (at a displacement)
– M = 10 kg
– Vmax = [2(PE)/M]1/2 = 1.4m/s (at zero displacement)
Energy is Conserved:
E = KE (is0) + PE ( x
1
2
2
= Init ) = 10 J
2
E = KE ( @ x = 0 ) + 0(isPE ) = 12 mvmax
2
vmax
2
2
⎛ m 5⎞
2
= ⋅ 10m = ⋅ 10 = 2 ⎜
⎟
10
sec
m
⎝
⎠
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First Law of Thermodynamics
First Law: Energy of the Universe is Constant
E = q + w (remember this)
q = heat. Transferred between two bodies of differing
temperature. Note: q ≠ Temp!
w = work (transferred). Force acting over a distance (F x d)
(do not do work if something does not move)
Energy (of the system) can be found by the state of the system;
q and w depend to the transfer and cannot be deduced from
the state of the system.
The change in PE is the work done. Force is mg and distance
is h or z.
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Applying the First Law
• Need to differentiate
between the system and
surroundings.
Surroundings
System
q transfer
• System: That part of the
universe you are interested
in (i.e., you define it).
• Surroundings: The rest of
the universe.
w transfer
The change (DELTA) in energy
of the system
ΔE ≡ ΔEsystem
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Conservation of Energy
• Total energy is conserved.
P = 1atm
• Energy gained by the
system must be lost by the
surroundings.
Initial
• Energy exchange can be in
the form of q, w, or both.
• How much heat was
withdrawn to cool the
system? Great Question,
we will answer it now.
P = 1atm
Final
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Heat Exchange: Exothermic
Ene rgy
Water @ 80° C
E initial
Water @ 20° C
q
E final
E final < E initial
• Exothermic Reaction.
Chemical process in which
system evolves resulting in
heat transfer to the
surroundings.
• q < 0 (heat is lost)
• Heat goes out of system, so
energy of system drops.
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Another Example of Exothermic
The loss in potential energy within the system results in a release of
heat into the environment (q<0), no work (w=0) Thermite Reaction!
ΔE = ΔKE + ΔPE = q + w = q < 0
ΔKE = 0 (isothermal)
ΔE = q < 0
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Heat Exchange: Endothermic
Ene rgy
Water @ 80° C
E final
Water @ 20° C
q
• Endothermic Reaction:
Chemical process in
which system evolves
resulting in heat
transfer to the system.
E initial
E final > E initial
• q > 0 (heat is gained)
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Another Example of Endothermic
NO is unstable w.r.t. air. So why do we get any NO from our cars? This
question presages what is to come: There is more to chemical equilibrium that
just whether the reaction is endo- or exo- thermic!!!
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Energy and Sign Convention
Ene rgy
E initial
Eout
E final
ΔE < 0
Ene rgy
E final
E initial
Ein
• If system loses energy:
Efinal < Einitial
Efinal-Einitial = ΔE < 0.
• If system gains energy:
Efinal > Einitial
Efinal-Einitial = ΔE > 0.
ΔE > 0
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Heat and Work Sign Convention
ΔE = q + w < 0
If system gives up heat
q < 0 (q is negative)
If system gets heat
q > 0 (q is positive)
If system does work
w < 0 (w is negative)
If work is done on system
w > 0 (w is positive)
ΔE = q + w > 0
Which of these two cases best describes the thermite reaction?
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Problem Z9.20
•
A mixture of gas and air (in a cylinder, 50 cc) The explosion releases
1kJ of energy. What is the final volume of the new product (water and
CO2) expanding against one Atmosphere (or 10^5 Pascals). Assume
all of the energy is converted to work to expand the cylinder.
•
How to solve:
–
–
–
–
Know the energy released, and that goes into work only (so q=0)
Know that work and volume are related: w = − PdV
First law: ΔE = q + w = 0 − PΔV
Energy is released. This is important, it tells you the system gives up this
much energy, so the change in energy is negative.
ΔE = −1 ⋅ 103 J = − PΔV = −1 ⋅ 105 ⋅ ΔV
ΔV = +10−2 m 3 = 10 = 104 cc
ΔV = VF − VI = 104 cm 3
VF = VI + 104 = 50 + 10,000 ~ 10,000cm 3
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Example: Piston Expansion
• Figure 9.4, expansion
against a constant external
pressure
• No heat exchange:
q = 0 (adiabatic)
• The amount of work the
System does is negative:
w<0
Because the system is doing
the work, expanding the
gas.
Pext < P
ΔV = Vf −Vi = Ahf − Ahi = AΔh
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Example (cont.)
• How much work does the system do?
• Pext = force/area= mass*g/area
• w = force * distance = (m*g)*Δh
= -Pext * A * Δh
= -Pext * ΔV
• w = - Pext ΔV (note sign)
• When will it stop expanding?
Z9.10 Why the negative sign on work?
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Example 9.1, and Problem Z9.19
• A hot-air balloon is inflated from 4 * 106 l to 4.5 x
106 l by the addition of 1.3 x 108 J of heat. If the
balloon expands against an external pressure of 1
atm, what is ΔE for this process?
– (FYI) A sphere 20 meters in diameter has a volume
about 4 * 106 l . (just over 20 yards; c.f. football field).
• Solve:
– First, define the system: the balloon and the air inside
the balloon.
– Key: Some of the heat energy (q) will go into work (w),
and the rest goes into internal energy of the gas in the
balloon.
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Example 9.1 (cont.)
ΔE = q + w
= (1.3 x 108 J) + (-PΔV)
w= (-1 atm (Vfinal − Vinit))
= (-0.5 x 106 l.atm)
1 − Atm = 1.01 ⋅ 105 ⋅ 10−3 Pa ⋅ m 3 ∼ 102 J
• Conversion: 101.3 J per l.atm
w=(-0.5 x 106 l.atm) x (101.3 J/l.atm) = -5.1 x 107 J
The SI unit of pressure is the Pascal. 1 J is a Pa*m3
1 bar is also an SI unit and is exactly 1 x 105 Pa, and
1.013 bars is 1 Atmosphere.
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Example 9.1 (cont.)
ΔE = (13 x 107 J) + (-5 x 107 J)
= 8 x 107 J (Ans.)
In English: the system gained more energy
through heat than it lost doing work; which
had better be the case or it would not
expand on heating. Therefore, the overall
energy of the system has increased.
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