Transcript HEAT, WORK AND INTERNAL ENERGY - Weebly

HEAT, WORK AND
INTERNAL ENERGY
GLOBAL
WARMING?
THERMODYNAMICS: the science of energy,
specifically heat and work, and how the transfer of
energy effects the properties of materials.
Thermodynamics:
"thermo": Greek therme heat
"dynamics": Greek dynamikos
powerful
Physics that deals with the mechanical
action or relations between heat and
work
Example 1: Heat to work
Heat Q from flame provides energy
to do work
--------------------------------------------------------Example 2: Work to heat.
Work done by person is converted
to heat energy via friction.
A “system” is the “collection of objects on which
attention is being focused”
The “surroundings” are everything else in the
environment
The system and surroundings must be separated by
walls which can either insulate or allow heat flow
OPEN SYSTEM: Mass and energy freely moves in
and out between the system and the surrounding
ISOLATED SYSTEM: No interaction between the
system and the surrounding
CLOSED SYSTEM: fixed mass
Heat, Q, energy caused by
temperature difference
Heat
... is the amount of internal energy
entering or leaving a system
... occurs by conduction, convection,
or radiation.
... causes a substance's temperature
to change
... is not the same as the internal
energy of a substance
... is positive if thermal energy flows
into the substance
... is negative if thermal energy flows
out of the substance
... is measured in joules
Thermal Equilibrium
Systems (or objects) are said to be in thermal
equilibrium if there is no net flow of thermal energy
from one to the other. A thermometer is in thermal
equilibrium with the medium whose temperature it
measures, for example.
If two objects are in thermal equilibrium, they are at
the same temperature.
Work, W, energy caused by physical
motion
WORK
W is positive if work is
done by system.
W is negative if work is done on the
system.
Air does work on the
environment: W > 0.
Environment (man) does work on
system: W < 0
(Alternative: system does negative
work because force by air pressure
on thumb is opposite to the direction
of motion of the thumb.)
RELATIONSHIP BETWEEN HEAT AND WORK
Why does the volume of gas expands when it is heated?
W=Fxd
Pressure (P) = (Force) F
or
F=PA
(Area) A
Volume (V) = L x W x H or A x d
d=V
A
W=PAV =PV
A
Internal Energy (U or E)
: (measured in joules)
- Sum of random
translational, rotational,
and vibrational kinetic
energies
DU: change in U
DU > 0 is a gain of
internal energy
DU < 0 is a loss of
internal energy
---------------------------------Thermal Energy:
same as internal
energy
Vibrational kinetic
energy in solids.
The hotter the
object, the larger
the vibrational
kinetic energy
Motions of a
diatomic
molecule in a
fluid
INTERNAL ENERGY (U or E)
is the total of the kinetic energy due to the motion of
molecules (translational, rotational, vibrational) and the
potential energy associated with the vibrational and
electric energy of atoms within molecules or crystals.
The First Law of Thermodynamics states that :
The internal energy of a system changes
from an initial value Ui to a final value Uf due to
heat added (Q) and work done by the system
(W)
DU = Uf – Ui = Q – W

Q is positive when the system gains heat, and

negative when the system loses heat.
W is positive when it is done BY the system, and
negative when it is done ON the system
Example: 1000 J of thermal energy
flows into a system (Q = 1000
J). At the same time, 400 J of work
is done by the system (W = 400 J).
What is the change in the system's
internal energy U?
----------------------------------------------------------
Solution:
DU = Q - W
= 1000 J - 400 J
= 600 J
Example: 800 J of work is done on a
system (W = -800 J) as 500 J of
thermal energy is removed from the
system (Q = -500 J).
What is the change in the system's
internal energy U?
-----------------------------------------------------
Solution:
DU = Q - W
= -500 J - (-800 J)
= -500 J + 800 J
= 300 J
Work Done by an Expanding Gas
W = PDV
DV = Vf - Vi
W = P (Vf - Vi)
Area under pressure-volume
curve is the work done
----------------------------------------Isobaric Process: "same
pressure"
Greek: barys, heavy
Work and the Pressure-Volume Curve
Work Done = Area Under PV
curve
------------------------------------How much work is done by the
system when the system is
taken from:
(a) A to B (900 J)
(b) B to C (0 J)
(c) C to A (-1500 J)
------------------------------------Each "rectangle" has an area of
100 Pa-m3 = 100 (N/m2)-m3
= 100 N-m
= 100 Joules
Expanding Gas
10 grams of steam at 100oC at
constant pressure rises to
110oC:
P = 4 x 105 Pa
DT = 10oC
DV = 30.0 x 10-6 m3
c = 2.01 J/g oC
Example: If a gas
expands at a
constant pressure,
the work done by the
gas is:
W = PDV
What is the change in internal
energy?
DU = Q - W
W = (4 x 105)(30.0 x 10-6) = 12 J
Q = mcDT = (10)(2.01)(10) = 201 J
DU = Q - W = 201 J - 12 J = 189 J
Work, Rubber Bands, and Internal Energy
DU = Q - W
Expand rubber band:
W < 0, Q = 0 DU >0
temperature increases
------------------------------------------Press thick rubber band to
forehead and expand
it rapidly. The warming should
be obvious.
Now allow the band to
contract quickly;
cooling will also be evident.
ISOTHERMAL-Temperature remains constant
ISOBARIC - Pressure remains constant
ISOMETRIC - Volume remains constant
(also ISOVOLUMETRIC or ISOCHORIC)
Since ΔV = 0, W = 0 then DU = Q - W = Q
Adiabatic Expansion of a Ideal Gas
No heat transfer therefore no temperature change (Q=0).
Generally obtained by surrounding the entire system
with a strongly insulating material or by carrying out the
process so quickly that there is no time for a significant
heat transfer to take place.
If Q = 0 then ΔU = - W
A system that expands under
adiabatic conditions does
positive work, so the internal
energy decreases.
A system that contracts
under adiabatic conditions
does negative work, so the
internal energy increases.
Adiabatic Expansion of a Ideal Gas
Both adiabatic expansion and
compression of gases occur in
only hundredths of a second in
the cylinders of a car’s engine.
Blowing air through wide open mouth
results to warm air. Blowing through
small opening results to cooler air due
to adiabatic expansion.
Compresses air leaking out through
a small opening also results in
adiabatic cooling.
PROCESS DIAGRAMS: visualize
processes using properties (T, P, V, etc.)
Area underneath the
slope represents the
amount of work
done (P x V).
CYCLE: a system undergoes processes returning to its initial state
Area
underneath
the slope
represents
the amount
of work done
(P x V).
Refrigerators work by taking heat from the interior and
depositing it on the exterior
 The compressor raises the pressure and temperature
of the refrigerant (freon or ammonia) while the coils
OUTSIDE the refrigerator allow the now hot
refrigerant to dissipate the heat
 The warm refrigerant flows through an expansion
valve from a high-pressure to a low-pressure zone, so it
expands and evaporates
• The coils INSIDE the
refrigerator allow the cold
refrigerant to absorb heat,
cooling the interior
• The cool refrigerant flows
back to the compressor, and
the cycle repeats

Second Law of Thermodynamics
Heat flows naturally from a
region at high temperature to
a region at low
temperature. By itself, heat
will not flow from a cold to a
hot body.
When an isolated system
undergoes a change, passing
from one state to another, it
will do so in such a way that
its entropy (disorder) will
increase, or at best remain the
same.
ENTROPY
Can you beat the Second Law?




So, can you cool your kitchen by
leaving the refrigerator door open
NO!
The heat removed from the interior
of the refrigerator is deposited back
into the kitchen by the coils on the
back!
And to make matters worse, the Second Law of
Thermodynamics says that work is needed to move the
heat from cold to hot, so the actual amount of heat
added to the kitchen is MORE than the amount
removed from the refrigerator
Hopefully, you understand today’s lesson.
Otherwise, you’ll end up like this cow.