The Second Law of Thermodynamics Physics 102 Professor Lee Carkner Lecture 6

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 )  W = (2493)(2-1) = (2493) (1) = 2493 J  Step 2: Isochoric temperature drop to 300 K  No volume change = no work  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  Since  T = 0,  U = 0  Total heat   U = Q – W  Q =  U + W = = J  Heat is positive since heat flows in  To balance work that goes out

Engines   General engine properties:  A working substance (usually a gas)   An output of work (W)  An output of heat to a cool reservoir (Q L )

Papin’s Device

Heat and Work Over the Cycle   Since the net heat is Q H -Q L, from the first law of thermodynamics:  W = Q H - Q L  e = W/Q H  Can also write as: e = 1 - (Q L /Q H )

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   You cannot completely eliminate friction, turbulence etc.  The first law of thermodynamics says:   The second law of thermodynamics says:  You cannot break even

Carnot Engine  e ideal = 1 - (T L / T H )  This is the Carnot efficiency (or ideal efficiency)   Any engine operating between two temperatures is less efficient than the Carnot efficiency e < e ideal  There is a limit as to how efficient you can make your engine

Dealing With Engines  W = Q H - Q L e = W/Q H = (Q H - Q L )/Q H = 1 - (Q L /Q H )  Efficiency must be less than or equal to e ideal e < e ideal = 1 - (T L /T H )   Note that heats are absolute values and T must be in Kelvin  For individual parts of the cycle you can often use the ideal gas law (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 L 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 QLQL QHQH

Refrigerator Performance  Input equals output: Q L + W = Q H  COP = Q L / W COP = Q L /(Q H -Q L )  Unlike efficiency, COP can be greater than 1  COP for real refrigerators ~ 5  COP ideal = T L /(T H -T L )   This is the maximum COP for a fridge operating between these two temperatures (COP < COP ideal )

Refrigerators and the Second Law  You cannot move heat from low to high temperature without the addition of work   COP cannot be infinite  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  Exam Friday

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