Chapter 15 - The Laws of Thermodynamics

**SUMMARY**

**Thermodynamics** describes the processes whereby *energy is transferred*
as heat and work, *e.g.*, in steam engines, electric power generating
plants, energy is transferred from coal (heat as it is burned) to electrical
energy which can do useful work.

**First Law of Thermodynamics**

DU = Q - W

The change in internal energy, DU is equal
to the heat *added* minus work done *by* the system. (Q is *positive*
when heat is *added* to the system, and work, W, is *positive*
when work is done *by* the system; Otherwise they are negative.) The
first law is a statement of energy conservation.

**Isothermal process** means temperature, T, is constant. Since internal
energy is proportional to temperature, the change in internal energy, DU,
for an isothermal process is zero. Thus Q = W.

**Adiabatic** process means *no heat* is added or given off,
*i.e.*, Q = 0.

**Work done** by simple system *e.g.*, a piston undergoes expansion
or contraction.

Work W = PDV

where P is constant and DV = V_{f}
- V_{i} is change in volume. This is an isobaric process since
the pressure is constant.

**Second Law of Thermodynamics** - Can be stated in many different
forms.

1. Heat flows naturally from a hot object to a cold object and not vice
versa.

2. It is impossible to get only work from heat.

3. The entropy of a system and its surrounding always increases as a result
of natural process.

Define

**Efficiency** e = W/Q_{H}

where Q_{H} is heat that flows from the hot reservoir. Q_{L}
= Q_{H} - W is heat discharged to the cold (or low temperature)
reservoir. (Efficiency is what you get divided by what you put in.)

Thus e = 1 - Q_{L}/Q_{H}

**Carnot cycle** (ideal or maximum) efficiency

e_{ideal} = 1 - T_{L}/T_{H}

where T_{L} and T_{H} are the temperatures of the cold
and hot reservoirs respectively. The Carnot efficiency describes the maximum
efficiency possible. It is an ideal system. No real systems can have greater
than this efficiency. It is always less.

**Refrigerators, air conditioners and heat pumps** Coefficient of
performance - is a measure of how efficient the refrigerator is. The higher
the number, the better the performance. For an ideal system,

Where T_{L} is the inside temperature and T_{H} is the
outside temperature.

**Change in entropy **DS = Q/T

where Q is the heat added or removed and T is the temperature in Kelvin. Entropy is a measure of order or disorder of the system. Restatement of second law of thermodynamics - Natural processes tend to move toward a state of greater disorder. (Just look at the mess in your room!)

**Homework solutions**

3. Sketch a PV diagram.

Before you can sketch the PV diagram you have to find all the P and V values.

State 1: V_{1} = 2.0 L, P_{1} = 1 atm

State 2: V_{2} = 1.0 L, P_{2} = 1 atm

State 3: Isothermal expansion V_{3} = 2.0 L, by ideal gas law
P_{3} = (P_{2}V_{2})/V_{3} = 0.5 atm

State 4 same as state 1: P_{4} = P_{1} and V_{4}
= V_{1} = 2.0 L

9. Two-step process as shown on the figure below. Heat loss going from
a to b at constant volume, and heat added going from b to c where the temperature
at c is the same as in a. Calculate, (a) total work, change in internal
energy of the gas, and (c) total heat flow into or out of the gas. (Conversion
factors 1 atm = 1.0135´10^{5}
N/m^{2}, and 1 L = 10^{-3} m^{3}.)

a. Going from a to b, the work done is zero since W = DV and the volume is constant. However, going from b to c the work done is

W_{bc} = PDV_{bc}

= (1.5 atm)(1.0135´10^{5} N/m^{2}×atm)(10.0
L - 6.8 L)´10^{-3} m^{3}/L
= 486 J.

Total work W = 486 J.

b. The change in internal energy is given by

DU = ^{3}/_{2}nR DT
= ^{3}/_{2}nR(T_{c} - T_{a})

But given T_{c} = T_{a}. Thus DU
= 0. (This does not mean that the change in internal energy along ab or
along bc is zero. One is the negative of the other.)

c. The heat flow can be derived from the first law of thermodynamics.

DU = Q - W

Since DU = 0, Q = W = 486 J.

14. Calculate the metabolic rate for a 24-h activity.

Activity |
Calorie count |
Total |

Sleeping for 8 h | 60 kcal/h´8 h = | 480 kcal |

Desk for 8 h | 100 kcal/h´8 h = | 800 kcal |

Light work for 4 h | 200 kcal/h´4 h = | 800 kcal |

TV for 2 h | 100 kcal/h´2 h = | 200 kcal |

Tennis for 1.5 h | 400 kcal/h´1.5 h = | 600 kcal |

Run for 0.5 h | 1000 kcal/h´05 h = | 500 kcal |

Total = | 3380 kcal |

Thus the metabolic rate for a 24 h period is 3380 kcal.

17. Engine does 7200 J of work in each cycle while absorbing 12.0 kcal
from a high temperature reservoir. What is the efficiency?

Q_{H} = 12.0 kcal = (12.0´10^{3})(4.186
J)

= 50,232 J

Efficiency
*i.e.*, output over energy put in

= 7200 J/50,232 = 0.143 = 14.3%.

20. Nuclear power plant operates at 75% of its maximum
theoretical efficiency between T_{H} = 600°C
and T_{L} = 350°C. Plant produces
electric energy at 1.1 GW (*i.e.*, 10^{9} W), how much exhaust
heat is discharge per hour?

T_{H} = 600°C + 273 = 873 K

T_{L} = 350°C + 272 = 623 K (Temperature
must be in Kelvin)

Maximum efficiency for Carnot cycle is

However, the plant is operating at 75% of the maximum possible efficiency. This gives the operating efficiency to be

(0.286)(0.75) = 0.215

Basic definition of efficiency e = W/Q_{H}

or Q_{H} = W/e

= 1.1´10^{9} J/0.215

= 5.12´10^{9} J

Q_{H} is the total heat input of which 1.1´10^{9}
J comes out as work. Thus

Q_{H} = 5.12´10^{9}
J - 1.1´10^{9} J

= 4.01´10^{9} J is the amount
of heat discharged per second.

In 1 h total heat discharge = (4.02´10^{9}
J)(3600 s/h)

= 1.45´10^{13} J.

23. Heat engine utilizes a heat source at 550°C
with an ideal efficiency of 30%. Find new high temperature if the efficiency
is to be increased to 40%.

For an ideal Carnot engine, the efficiency is

For a 30% efficient engine, 0.3 = 1 - T_{L}/(273 + 550)K

Solving gives T_{L} = 576 K

Now for a 40% efficient engine, 0.4 = 1 - 576 K/T_{H}

The new high temperature is T_{H} = 960 K = 687°C.

Note: To use any of the formula in this chapter the temperature **must
be in Kelvin**.

28. A restaurant refrigerator has coefficient of performance, COP = 5.0. (COP is a measure of how efficient the refrigerator is. The higher the number, the more efficient it is.)

For refrigerators,

Solving gives T_{L} = 252 K = -21°C.

33. 1 kilogram of H_{2}0 heated from 0°C
to 100°C. Estimate the change in entropy
of H_{2}0.

From 0°C to 100°C

Heat Q = mcT

= (1 kg)(4186 J/kg×C°)(100°C - 0°C)

= 4.186´10^{5} J.

Entropy change DS = Q/T

Since temperature goes from 0°C to 100°C, take average,

T_{avg} = 50°C = 50°C
+ 273K = 323 K

Hence

38. A 5.0 kg piece of Al at 30°C placed
in 1.0 kg of water in a Styrofoam container at room temperature (20°C
= 293 K). Calculate the approximate net change in entropy of the system.
Need the equilibrium temperature.

Heat lost by Al = heat gained by water

m_{Al}c_{Al}DT_{Al}
= m_{w}c_{w}DT_{w}

(5.0 kg)(900 J/kg×C°)(30°C
- T_{c}) = (1.0 kg)(4186 J/kg×C°)(T_{c}
- 20°C)

Solving gives T_{c} = 25.24°C.

Thus heat lost by Al, Q_{Al} = m_{Al}c_{Al}DT_{Al}
= 2.142´10^{4} J.

Average temperature of aluminum = (30 + 25.24)/2 = 27.629°C = 300.62 K.

Hence entropy change for aluminum,

= -70.25 J/K

Average temperature of water is (20 + 25.24)/2 = 22.62°C = 295.62 K

And entropy change for water,

= +72.46 J/K

Hence net entropy change is DS_{net}
= DS_{Al} + DS_{w}

= 1.21 J/K

The net entropy change is greater than or equal to 0 as required by
the 2^{nd} law of thermodynamics.

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This page last updated on September 19, 1997.

© 1996 Dr. H. K. Ng.

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