Laws of Thermodynamics

Laws of Thermodynamics Thermal Physics, Lecture 4

Internal Energy, U • Internal Energy is the total energy in a substance, including thermal, chemical potential, nuclear, electrical, etc. • Thermal Energy is that portion of the internal energy that changes when the temperature changes

Heat, Q • The flow of energy into or out of a substance due to a difference in temperature. • It results in a loss or gain in the Thermal Energy

Work, W • The flow of energy into or out of a substance that is NOT due to a difference in temperature. • It results in a loss or gain in the Internal Energy

First Law • The first law of thermodynamics states that the internal energy of a system is conserved. • Q is the heat that is added to the system • If heat is lost, Q is negative. • W is the work done by the system. • If work is done on the system, W is negative

Example using First Law • 2500 J of heat is added to a system, and 1800 J of work is done on the system. What is the change in the internal energy of this system? • Signs: Q=+2500 J, W= -1800 J

Thermodynamic Processes • We will consider a system where an ideal gas is contained in a cylinder fitted with a movable piston

Isothermal Processes • Consider an isothermal process iso = same, so isothermal process happens at the same temperature. • Since PV=nRT, if n and T are constant then PV = constant

Isothermal Processes • We can plot the pressure and volume of this gas on a PV diagram • The lines of constant PV are called isotherm’s

Isothermal Processes • For an ideal gas, the internal energy U depends only on T, so the internal energy does not change. • Q must be added to increase the pressure, but the volume expands and does work on the environment. • Therefore, Q=W

Adiabatic Processes • In an adiabatic process, no heat is allowed to flow into or out of the system. • Q=0 • Examples: • well-insulated systems are adiabatic • very rapid processes, like the expansion of a gas in combustion, don’t allow time for heat to flow (Heat transfers relatively slowly)

Adiabatic Processes

Isobaric Processes • Isobaric processes happen when the pressure is constant

Isovolumetric Processes • Isovolumetric processes happen when the volume is constant

Calculating Work • For our ideal gas undergoing an isobaric process

Calculating Work • What if the pressure is not constant? • The work is the area under the curve of the PV diagram.

Adiabatic Process • Stretch a rubber band suddenly and use your lips to gauge the temperature before and after.

Summary of Processes • Isothermal: T is constant and Q=W since U=0 • Isobaric: P is constant and W=PV • Isovolumetric: V is constant so W=0 and Q = U • Adiabatic: Q=0 so U = -W

Example • An ideal gas is slowly compressed at constant pressure of 2 atm from 10 L to 2 L. Heat is then added to the gas at constant volume until the original temperature is reached. What is the total work done on the gas?

Example

Example (15-5 in textbook) • Work is area under the graph. • Convert pressure to Pa and Volume to cubic meters.

Example (15-5 in textbook) • How much heat flows into the gas?

Example 2: Boiling Water • 1 kg (1 L) of water at 100 C is boiled away at 1 atm of pressure. This results in 1671 L of steam. Find the change in internal energy.

Example 2: Boiling Water

HEAT ENGINE REFRIGERATOR system TH TH QH QH W W QC QC TC TC Engines and Refrigerators • system taken in closed cycle  Usystem = 0 • therefore, net heat absorbed = work done QH - QC = W (engine) QC - QH = -W (refrigerator)

HEAT ENGINE TH QH W QC TC Heat Engine: Efficiency The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: QH -QC = W efficiency e  W/QH

Heat Engine • 1500 J of energy, in the form of heat, goes into an engine, which is able to do a total of 300 J of work. What is the efficiency of this engine? • What happens to the rest of the energy?

HEAT ENGINE TH 300K QH W QC TC = 77K Heat Engine • Can you get “work” out of a heat engine, if the hottest thing you have is at room temperature? • A) Yes B) No

REFRIGERATOR TH QH W QC TC Refrigerator The objective: remove heat from cold reservoir The cost: work 1st Law: QH = W + QC coeff of performance Kr QC/W

New concept: Entropy (S) • A measure of “disorder” • A property of a system (just like p, V, T, U) related to number of number of different “states” of system • Examples of increasing entropy: • ice cube melts • gases expand into vacuum • Change in entropy: • S = Q/T • >0 if heat flows into system (Q>0) • <0 if heat flows out of system (Q<0)

Second Law of Thermodynamics • The entropy change (Q/T) of the system+environment is always greater than zero (positive) • never < 0 • Result: order to disorder • Consequences • A “disordered” state cannot spontaneously transform into an “ordered” state • No engine operating between two reservoirs can be more efficient than one that produces 0 change in entropy. This is called a “Carnot engine”

Carnot Cycle • Idealized (Perfect) Heat Engine • No Friction, so DS = Q/T = 0 • Reversible Process • Isothermal Expansion • Adiabatic Expansion • Isothermal Compression • Adiabatic Compression

Perpetual Motion Machines?

Carnot Efficiency • The absolute best a heat engine can do is given by the Carnot efficiency:

Carnot Efficiency • A steam engine operates at a temperature of 500 C in an environment where the surrounding temperature is 20 C. What is the maximum (ideal) efficiency of this engine? • What is the operating temperature were increased to 800 C ?

HEAT ENGINE TH QH W QC TC Engines and the 2nd Law The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: QH -QC = W efficiency e  W/QH =W/QH = 1-QC/QH

Summary • First Law of thermodynamics: Energy Conservation • Q = DU + W • Heat Engines • Efficiency = 1-QC/QH • Refrigerators • Coefficient of Performance = QC/(QH - QC) • Entropy DS = Q/T • 2nd Law: Entropy always increases! • Carnot Cycle: Reversible, Maximum Efficiency e = 1 – Tc/Th

Laws of Thermodynamics