# Entropy and the Second Law of Thermodynamics

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Entropy and the Second Law of Thermodynamics

Heat Engine and Refrigerators
A heat engine is a device that carries a working substance through a cyclic process, during which it Absorbs thermal energy from a high-temperature source (H) The engine does work and Expels thermal energy to a low-temperature source (L)

Thermal efficiency: 1st Law Cyclic Process Work done by the substance
Heat added to the substance Thermal efficiency:

A refrigerator is a heat engine in reverse.
A refrigerator is a device that carries a working substance through a cyclic process, during which it Absorbs thermal energy from a low-temperature source by doing Work Expels thermal energy (the amount work done and the heat absorbed) to a high-temperature source

Coefficient of performance:
1st Law Cyclic Process Heat absorbed by the substance Work done on the substance Coefficient of performance:

Reversible and Irreversible Processes
Reversible Process: a process that at the conclusion, the system and its surrounding return to the exact initial conditions. All natural processes are irreversible, at best “almost” reversible.

Irreversible Processes
Heat flows from the high-temperature to the low-temperature objects. The bouncing ball finally stops The oscillating pendulum finally stops

Carnot Engine (an Ideal Engine)
Carnot engine: a heat engine operating in a Carnot cycle. Carnot cycle: a cycle consisting of four REVERSIBLE (therefore Ideal) processes: two adiabatic processes B C; D A and two isothermal processes. A B;C D Carnot Carnot engine is the most efficient heat engine possible.

Efficiency of a Carnot Engine
Isothermal Isothermal: A  B C  D Adiabatic Adiabatic: B  C D  A

Efficiency of an Ideal Carnot Engine
Thermal efficiency is then

Performance of an Ideal Carnot Refrigerator
Coefficient of performance

HRW 38E (5th ed.). An ideal heat engine operates in a Carnot cycle between 235˚C and 115˚C. It absorbs 6.30x104 J per cycle at the higher temperature. (a) What is the efficiency of the engine? (b) How much work per cycle is this engine capable of performing? (a) e = T H - C ( 235 115 ) K + 273 23 . 6 % (a)

HRW 45P (5th ed.). One mole of an ideal monatomic gas is taken through the cycle shown. Assume that p =2p0, V =2V0, p0 = 1.01x105 Pa, and V0 = m3. Calculate (a) the work done during the cycle, (b) the heat added during stroke abc, and (c) the efficiency of the cycle. (d) What is the efficiency of an ideal engine operating between the highest and lowest temperatures that occur in the cycle? How does this compare to the efficiency calculated in (c)? a c d Volume Pressure b V0, p0 V, p (a) (b)

HRW 45P (5th ed.). One mole of an ideal monatomic gas is taken through the cycle shown. Assume that p =2p0, V =2V0, p0 = 1.01x105 Pa, and V0 = m3. Calculate (a) the work done during the cycle, (b) the heat added during stroke abc, and (c) the efficiency of the cycle. (d) What is the efficiency of an ideal engine operating between the highest and lowest temperatures that occur in the cycle? How does this compare to the efficiency calculated in (c)? a c d Volume Pressure b V0, p0 V, p (b) (c)

HRW 45P (5th ed.). One mole of an ideal monatomic gas is taken through the cycle shown. Assume that p =2p0, V =2V0, p0 = 1.01x105 Pa, and V0 = m3. Calculate (a) the work done during the cycle, (b) the heat added during stroke abc, and (c) the efficiency of the cycle. (d) What is the efficiency of an ideal engine operating between the highest and lowest temperatures that occur in the cycle? How does this compare to the efficiency calculated in (c)? a c d Volume Pressure b V0, p0 V, p (d) Larger than 15%.

For a complete cycle, ∆Eint = 0
HRW 60P (5th ed.). An ideal heat pump is used to heat a building. The outside temperature is -5.0˚C and the temperature inside the building is to be maintained at 22˚C. The coefficient of performance is 3.8 and the heat pump delivers 7.54 MJ of heat to the building each hour. At what rate must work be done to run the heat pump? For a complete cycle, ∆Eint = 0 In one hour Rate = 1.57 x 106 J/3600 s = 440 W

Entropy and the Second Law of Thermodynamics

The Second Law of Thermodynamics
It is impossible to construct a heat engine with 100% efficiency. It is impossible to construct a refrigerator that does not require work.

Absolute Temperature Scale
Carnot engine — the most efficient heat engine: TC = 0  e = 1 Second Law  Temperature cannot be equal or lower than T = 0 K

Entropy Entropy is a state function like p, T, and Eint.
A state function describes the thermodynamic state of a system, which is independent of the history. The change in entropy S for an infinitesimal process: The subscript “r” stands for reversible process

For a finite process from an initial state “i” to a
final state “f”, the change in entropy is Unit: J/K The integration is along a path that represents a reversible process

From the statistical mechanical point of view, entropy
is a measure of disorder.

The Second Law of Thermodynamics
In a CLOSED system: No energy exchange with other systems = : reversible process > : irreversible process

Carnot Engine ò Ñ For the gas: the Carnot cycle is a reversible
Entropy Carnot Engine For the gas: the Carnot cycle is a reversible closed cycle — back to the same state: dQ ò D = S Ñ r = T

Carnot Engine For the hot reservoir: heat is lost A  B
Isothermal Carnot Engine For the hot reservoir: heat is lost A  B For the cold reservoir: heat is gained C  D

Carnot cycle Carnot Engine B  C and D  A: Adiabatic

HRW 5E (5th ed.). An ideal gas in contact with a constant-temperature reservoir undergoes a reversible isothermal expansion to twice its initial volume. Show that the reservoir’s change in entropy is independent of its temperature. Isothermal ∆Eint = 0  Q = W Change of entropy of the ideal gas: Reversible process: ∆Stotal = 0 Change of entropy of the reservoir: ∆S’ = -∆S = -nRln2

HRW 15P (5th ed.). A 2.0 mol sample of an ideal monatomic gas undergoes the reversible process shown. (a) How much heat is absorbed by the gas? (b) What is the change in the internal energy of the gas? (c) How much work is done by the gas? a. b. Monatomic Entropy (J/K) 5 10 Temperature (K) 200 400 15 20 c.

HRW 25P (5th ed.). A 10 g ice cube at -10˚C is placed in a lake whose temperature is 15˚C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg·K. The ice warms from -10˚C to 0˚C (1), then melts (2), then the water warms to the lake temperature of 15˚C (3). The entropy change of the ice: (1)

HRW 25P (5th ed.). A 10 g ice cube at -10˚C is placed in a lake whose temperature is 15˚C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg·K. (2) (3)

HRW 25P (5th ed.). A 10 g ice cube at -10˚C is placed in a lake whose temperature is 15˚C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg·K. The entropy change of the lake: assume its temperature does not change: ∆SL = Q/T (1) (2) (3)

HRW 25P (5th ed.). A 10 g ice cube at -10˚C is placed in a lake whose temperature is 15˚C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg·K.