Chapter 15 - The Laws of Thermodynamics
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 = Vf
- Vi is change in volume. This is an isobaric process since
the pressure is constant.
Second Law of Thermodynamics - Can be stated in many different
1. Heat flows naturally from a hot object to a cold object and not vice
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.
Efficiency e = W/QH
where QH is heat that flows from the hot reservoir. QL = QH - 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 - QL/QH
Carnot cycle (ideal or maximum) efficiency
eideal = 1 - TL/TH
where TL and TH 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 TL is the inside temperature and TH is the
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!)
3. Sketch a PV diagram.
Before you can sketch the PV diagram you have to find all the P and V values.
State 1: V1 = 2.0 L, P1 = 1 atm
State 2: V2 = 1.0 L, P2 = 1 atm
State 3: Isothermal expansion V3 = 2.0 L, by ideal gas law P3 = (P2V2)/V3 = 0.5 atm
State 4 same as state 1: P4 = P1 and V4 = V1 = 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´105
N/m2, and 1 L = 10-3 m3.)
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
Wbc = PDVbc
= (1.5 atm)(1.0135´105 N/m2×atm)(10.0 L - 6.8 L)´10-3 m3/L = 486 J.
Total work W = 486 J.
b. The change in internal energy is given by
DU = 3/2nR DT = 3/2nR(Tc - Ta)
But given Tc = Ta. 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.
|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?
QH = 12.0 kcal = (12.0´103)(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 TH = 600°C
and TL = 350°C. Plant produces
electric energy at 1.1 GW (i.e., 109 W), how much exhaust
heat is discharge per hour?
TH = 600°C + 273 = 873 K
TL = 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/QH
or QH = W/e
= 1.1´109 J/0.215
= 5.12´109 J
QH is the total heat input of which 1.1´109 J comes out as work. Thus
QH = 5.12´109 J - 1.1´109 J
= 4.01´109 J is the amount of heat discharged per second.
In 1 h total heat discharge = (4.02´109 J)(3600 s/h)
= 1.45´1013 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 - TL/(273 + 550)K
Solving gives TL = 576 K
Now for a 40% efficient engine, 0.4 = 1 - 576 K/TH
The new high temperature is TH = 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.)
Solving gives TL = 252 K = -21°C.
33. 1 kilogram of H20 heated from 0°C
to 100°C. Estimate the change in entropy
From 0°C to 100°C
Heat Q = mcT
= (1 kg)(4186 J/kg×C°)(100°C - 0°C)
= 4.186´105 J.
Entropy change DS = Q/T
Since temperature goes from 0°C to 100°C, take average,
Tavg = 50°C = 50°C + 273K = 323 K
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
mAlcAlDTAl = mwcwDTw
(5.0 kg)(900 J/kg×C°)(30°C - Tc) = (1.0 kg)(4186 J/kg×C°)(Tc - 20°C)
Solving gives Tc = 25.24°C.
Thus heat lost by Al, QAl = mAlcAlDTAl = 2.142´104 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 DSnet = DSAl + DSw
= 1.21 J/K
The net entropy change is greater than or equal to 0 as required by
the 2nd law of thermodynamics.
times since November 12, 1996.
This page last updated on September 19, 1997.
© 1996 Dr. H. K. Ng.
All Rights Reserved.