Post on 06-May-2015
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Thermodynamics
THERMODYNAMICSThermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy and heat transfer.
Central Heating
A THERMODYNAMIC SYSTEM• A system is a closed environment in which
heat transfer can take place. (For example, the gas, walls, and cylinder of an automobile engine.)
Work done on gas or work done by gas
INTERNAL ENERGY OF SYSTEM
• The internal energy U of a system is the total of all kinds of energy possessed by the particles that make up the system.
Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.
TWO WAYS TO INCREASE THE INTERNAL ENERGY, U.
HEAT PUT INTO A SYSTEM
(Positive)
+U
WORK DONE ON A GAS (Positive)
WORK DONE BY EXPANDING
GAS: W is positive
WORK DONE BY EXPANDING
GAS: W is positive
-UDecreas
e
-UDecreas
e
TWO WAYS TO DECREASE THE INTERNAL ENERGY, U.
HEAT LEAVES A SYSTEM Q is
negative
Qout
hot
WoutWout
hot
THERMODYNAMIC STATE
The STATE of a thermodynamic system is determined by four factors:
• Absolute Pressure P in Pascals
• Temperature T in Kelvins• Volume V in cubic
meters• Number of moles, n, of working gas
THE FIRST LAW OF THERMODYAMICS:
• The net heat put into a system is equal to the change in internal energy of the system plus the work done BY the system.
Q = U + W final - initial)
• Conversely, the work done ON a system is equal to the change in internal energy plus the heat lost in the process.
SIGN CONVENTIONS FOR FIRST LAW
• Heat Q input is positive
Q = U + W final - initial)
• Heat OUT is negative
• Work BY a gas is positive• Work ON a gas is negative
+Qin
+Wout
U
-Win
-Qout
U
APPLICATION OF FIRST LAW OF THERMODYNAMICS
Example 1: In the figure, the gas absorbs 400 J of heat and at the same time does 120 J of work on the piston. What is the change in internal energy of the system?
Q = U + W
Apply First Law:
Qin
400 J
Wout =120 J
Example 1 (Cont.): Apply First Law
U = +280 J
Qin
400 J
Wout =120 J
U = Q - W
= (+400 J) - (+120 J)
= +280 J
W is positive: +120 J (Work OUT)
Q = U + W
U = Q - W
Q is positive: +400 J (Heat IN)
FOUR THERMODYNAMIC PROCESSES:
• Isochoric Process: V = 0, W = 0 • Isobaric Process: P = 0 • Isothermal Process: T = 0, U = 0 • Adiabatic Process: Q = 0
• Isochoric Process: V = 0, W = 0 • Isobaric Process: P = 0 • Isothermal Process: T = 0, U = 0 • Adiabatic Process: Q = 0
Q = U + W
HEAT ENGINES
• Absorbs heat Qhot
• Performs work Wout
• Rejects heat Qcold
A heat engine is any device which through a cyclic process:
Cold Res. TC
Engine
Hot Res. TH
Qhot Wout
Qcold
THE SECOND LAW OF THERMODYNAMICS
It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.
Not only can you not win (1st law); you can’t even break even (2nd law)!
Wout
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
THE SECOND LAW OF THERMODYNAMICS
Cold Res. TC
Engine
Hot Res. TH
400 J
300 J
100 J
• A possible engine. • An IMPOSSIBLE engine.
Cold Res. TC
Engine
Hot Res. TH
400 J 400 J
EFFICIENCY OF AN ENGINE
Cold Res. TC
Engine
Hot Res. TH
QH W
QC
The efficiency of a heat engine is the ratio of the net work done W to the heat input QH.
e = 1 - QC
QH
e = = W
QH
QH- QC
QH
EFFICIENCY EXAMPLE
Cold Res. TC
Engine
Hot Res. TH
800 J W
600 J
An engine absorbs 800 J and wastes 600 J every cycle. What is the efficiency?
e = 1 - 600 J
800 J
e = 1 - QC
QH
e = 25%
Question: How many joules of work is done?
REFRIGERATORSA refrigerator is an engine operating in reverse: Work is done on gas extracting heat from cold reservoir and depositing heat into hot reservoir.Win + Qcold = Qhot
WIN = Qhot - Qcold
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
Win
THE SECOND LAW FOR REFRIGERATORS
It is impossible to construct a refrigerator that absorbs heat from a cold reservoir and deposits equal heat to a hot reservoir with W = 0.If this were possible, we could establish perpetual motion!
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold