Second The 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 m3 expands to 2 m3, constant pressure W = PV, PV = nRT P = nRT/V = (1)(8.31)(300)/(1) = 2493 Pa W = PV = (P) (VfVi) W = (2493)(21) = (2493) (1) = 2493 J Step 2: Isochoric temperature drop to 300 K No volume change = no work Total work = 2493 + 0 = +2493 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 = 0 + 2493 = +2493 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 (QL)
Papin's Device 1690
Heat and Work Over the Cycle
Since the net heat is QHQL, from the first law of thermodynamics: W = QH QL e = W/QH
Can also write as: e = 1 (QL /QH)
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 The first law of thermodynamics says:
You cannot break even
You cannot completely eliminate friction, turbulence etc.
The second law of thermodynamics says:
Carnot Engine
eideal = 1 (TL / TH)
This is the Carnot efficiency (or ideal efficiency)
Any engine operating between two temperatures is less efficient than the Carnot efficiency e < eideal There is a limit as to how efficient you can make your engine
Dealing With Engines
W = QH QL e = W/QH = (QH QL)/QH = 1 (QL/QH) Efficiency must be less than or equal to eideal e < eideal = 1 (TL/TH)
Note that heats absolute are 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 QL is input from the cold reservoir, W is input power, QH is output to the hot reservoir
How a Refrigerator Works
Compressor (work =W)
QL Heat removed from fridge by evaporation
Gas Low Pressure Liquid High Pressure
QH Heat added to room by condensation
Expansion Valve
Refrigerator Performance
Input equals output: QL + W = QH
COP = QL / W COP = QL/(QHQL) Unlike efficiency, COP can be greater than 1
COP for real refrigerators ~ 5
COPideal = TL /(THTL)
This is the maximum COP for a fridge operating between these two temperatures (COP < COPideal)
Refrigerators and the Second Law
You cannot move heat from low to high temperature without the addition of work
COP cannot be infinite You ...