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= 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

THTH TCTC QHQH QCQC W HEAT ENGINE THTH TCTC QHQH QCQC W REFRIGERATOR system l system taken in closed cycle   U system = 0 l therefore, net heat absorbed = work done Q H - Q C = W (engine) Q C - Q H = -W (refrigerator) Engines and Refrigerators

THTH TCTC QHQH QCQC W HEAT ENGINE The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: Q H -Q C = W efficiency e  W/Q H Heat Engine: Efficiency

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 Can you get “work” out of a heat engine, if the hottest thing you have is at room temperature? A) YesB) No T H 300K T C = 77K QHQH QCQC W HEAT ENGINE

THTH TCTC QHQH QCQC W REFRIGERATOR The objective: remove heat from cold reservoir The cost: work 1st Law: Q H = W + Q C coeff of performance K r  Q C /W Refrigerator

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  S = 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 ?

THTH TCTC QHQH QCQC W HEAT ENGINE The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: Q H -Q C = W efficiency e  W/Q H =W/Q H = 1- Q C /Q H Engines and the 2nd Law

Summary First Law of thermodynamics: Energy Conservation Q =  U + W Heat Engines Efficiency = 1-Q C /Q H Refrigerators Coefficient of Performance = Q C /(Q H - Q C ) Entropy  S = Q/T 2 nd Law: Entropy always increases! Carnot Cycle: Reversible, Maximum Efficiency e = 1 – T c /T h