The Second Law of Thermodynamics Physics 102 Professor Lee Carkner Lecture 7

PAL #6 First Law  Work of 2 step process  Step 1: 1 mole of gas at 300 K and 1 m 3 expands to 2 m 3, constant pressure  W = P  V, PV = nRT  P = nRT/V = (1)(8.31)(300)/(1) = 2493 Pa  W = P  V = (P) (V f -V i )   Step 2: Isochoric temperature drop to 300 K   Total work = = J  W is positive since gas is expanding and work is output

PAL #6 First Law  Change of internal energy  Start at 300 K, end at 300 K   Total heat   Q =  U + W = = J  Heat is positive since heat flows in 

Ten joules of heat are added to a cylinder of gas causing the piston at the top to rise. How much work does the piston do? A)0 Joules B)5 Joules C)10 Joules D)-10 joules E)You cannot tell from the information given

Which of the processes in the diagram produces the least work? A)1 B)2 C)3 D)4 E)All are the same

Which of the processes in the diagram has the least change in internal energy? A)1 B)2 C)3 D)4 E)All are the same

Which of the processes in the diagram involves the least heat? A)1 B)2 C)3 D)4 E)All are the same

Engines  An engine is a device for converting heat into work by continuously repeating a cyclic process  General engine properties:   An input of heat   An output of heat

Heat and Work Over the Cycle  Four parts of the cycle:   compression   output of heat Q C  Over the course of one cycle positive work is done and heat is transferred   Since the engine is a cycle, the change in internal energy is zero   U=(Q H -Q C )-W =0 W = Q H - Q C

Efficiency   In order for the engine to work we need a source of heat for Q H  Q H is what you put in, W is what you get out so the efficiency is: e = W/Q H   The rest is output as Q C  Note that all heats in engine problems are absolute values

Today’s PAL  If an automobile engine outputs W to the drive shaft and outputs W to the radiator, what is the efficiency?  If gas is $2.50 per gallon, how much money per gallon are you wasting?

Efficiency and Heat  e = 1 - (Q C /Q H )  The efficiency depends on how much of Q H is transformed into W and how much is lost in Q C : Q H = W + Q C 

The Second Law of Thermodynamics   This is one way of stating the second law: It is impossible to build an engine that converts heat completely into work   Engines get hot, they produce waste heat (Q C )  You cannot completely eliminate friction, turbulence etc.

Carnot Engine  In 1824 Sadi Carnot related the maximum efficiency to the temperature of the reservoirs: e C = 1 - (T C / T H )   A hot input reservoir and a cold output reservoir make it “easier” to move heat in and out  e < e C  There is a limit as to how efficient you can make your engine

The First and Second Laws  The first law of thermodynamics says:   The second law of thermodynamics says:   The two laws imply:   W < Q H   W  Q H

Dealing With Engines  Most engine problems can be solved by knowing how to express the efficiency and relate the work and heats: e = W/Q H = (Q H - Q C )/Q H = 1 - (Q C /Q H )  Efficiency must be less than or equal to the Carnot Efficiency:  If you know T C and T H you can find an upper limit for e (=W/Q H )  PV = nRT

Refrigerators   A refrigerator is a device that uses work to move heat from low to high temperature   The refrigerator is the device on the back of the box   Your kitchen is the hot reservoir  Heat Q C is input from the cold reservoir, W is input power, Q H is output to the hot reservoir

How a Refrigerator Works Liquid Gas Compressor (work =W) Expansion Valve Heat removed from fridge by evaporation Heat added to room by condensation High Pressure Low Pressure QCQC QHQH

Refrigerator Performance  Input equals output:  The equivalent of efficiency for a refrigerator is the coefficient of performance COP: COP = Q C / W  Unlike efficiency, COP can be greater than 1 

Refrigerators and Temperature  COP = T C /(T H -T C )   This is the maximum COP for a fridge operating between these two temperatures

Refrigerators and the Second Law  You cannot move heat from low to high temperature without the addition of work    You need to do work on the coolant in order for it to release the heat 

Next Time  Read:  Homework: Ch 15, P 26, 31, 35, 37  Practice problems now posted for Friday’s exam  Won’t count for grade