Chem 452B – HW 5 (B) Fall 2012
Homework 5B Key
Q1) An insulated water bath maintained at 273K contains 20 grams of ice. The pressure is
constant at 1 atm. A piece of nickel at a temperature of 373K is dropped into the bath.
The temperature of the bath does not change, but when the nickel cools to 273K, 10
grams of ice have melted. Calculate how much the nickel weighed, the entropy change of
the nickel, the entropy change of the bath+ice (i.e. surroundings) and the entropy change
of the universe. CP  Ni   0.46 J
CP  ice   2.09 J
H P  ice   334 J
gK
gK
g
The system consists of Ice Water and hot metal. This system is isolated from the environment.
Therefore: SEnv  0  STotal  0  SSys  S .
H  CP T  mICE H melt ,m  mNi  CP  Ni  P  TNi
0  0  10  334  mNi  0.46   100  J
10  334
 73.g
0.46  100 
The entropy change of the water is zero, the entropy change of the ice and water is just the
change of 10 grams of ice at 273 to10 grams of water at 273. The Ni change was just the
isothermal, reversible cooling from 373 to 273.
H ICE ,melt 10  334 J
S ICE 

 12.234 J K
Tmelt
273K
mNi  
273
S Ni 

373
C
P
 Ni  dT
T
C
P
 Ni  ln
273
 0.46  73  ln 0.732  10.48 J K
373
S Sys  S ICE  S Ni  12.234  10.48  1.75 J K  0
The total entropy change is positive; the process is spontaneous in the direction described. It is
not zero because the Ni was not changed reversibly. The two bodies had different initial
temperatures.
Q2) Text, 5.39 (Answers: (a) 0.627 (b)0.398 (c)~110 tons/hr) The Chalk Point, Maryland,
generating station supplies electrical power to the Washington, D.C., area. Units 1 and 2 have a
gross generating capacity of 710 megawatts (MW). The steam pressure is 25 106 Pa and the
super-heater outlet temperature (Th) is 540°C. The condensate temperature (Tc) is 30.0°C.
a. What is the efficiency of a reversible Carnot engine operating under these conditions?

Thi  Tlo
T
303
 1  lo  1 
 0.63
Thi
Thi
813
b. If the efficiency of the boiler is 91.2%, the overall efficiency of the turbine, which
includes the Carnot efficiency and its mechanical efficiency, is 46.7%, and the
efficiency of the generator is 98.4%, what is the efficiency of the total generating unit?
(Another 5% needs to be subtracted for other plant losses.
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Chem 452B – HW 5 (B) Fall 2012
The overall efficiency is 40%. 0.912*.467*.984*.95  0.40
c. One of the coal burning units produces 355 MW. How many metric tons (1 metric ton
= 1 106 g) of coal per hour are required to operate this unit at its peak output if the
enthalpy of combustion of coal is 29 103 kJ kg ?
C  O2  CO2
H  H rxn  X
X  m C
355 106 J
 m C 29 103 J g
sec
355 106 g
4 g
2 T
3 sec
T
m C 
s  1.22 10
s  1.22 10
s  3.6 10
hr  44 hr
3
29 10
This assumes 100% efficiency. If the overall efficiency is 40%, then the plant needs to burn
44
110 T hr  T hr .
.4
Q3) Find S for the following processes:
a. The isothermal reversible expansion at room temperature of one mole of an ideal gas
from 2L to 3L. (Answer: 3.4 J/K)
q
wrev
V
S  rev 
 nR ln 2  8.3  ln 32  3.4 J K
T
T
V1
b. One mole of ice at 265K is melted under usual lab conditions to form water at 329K.
(Answer: about 37 J/K)
g
H  ice   18 mol
 334 J
g
CP  ice   18  2.09 J
K
J
CP  water   75.37
K
Tmelt H melt
T
273
329
S  CP  ice  ln

 CP  water  ln
 37.6 ln
 24 J K  75.37 ln
T
Tmelt
Tmelt
265
273
S  1.1  24  14  39 J K
Q4) Two blocks of the same metal and mass are at different initial temperatures T1 and T2.
The blocks are brought into contact and come to a final temperature Tf. Assume the system is
totally isolated from the surroundings and that the surroundings are at equilibrium.
a. Your intuition probably tells you that for 2 identical blocks, Tf is the average of the
initial temperatures, so that Tf = 1/2 (T1 + T2). Show that for a system of two blocks
totally isolated from the surroundings that this is true.
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Chem 452B – HW 5 (B) Fall 2012
0  H  H1  H 2  CP 1 T1  CP  2  T2
f 
CP 1
CP 1  CP  2 
0  f T  T1   1  f T  T2 
T  fT1  1  f  T2 
f 
1
2
T
1
2
1
2
T1  T2    f
 12  T1  T2 
T1  T2 
b. Show that the change in entropy is S  CP ln
T1  T2 
2
4 T1 T2
CP 1 dT
C  2  dT
T
T
 P
 CP 1 ln  CP  2  ln
T
T
T1
T2
T1
T2
T
T
S  S1  S 2  
CP 1  CP  2   CP
T T 
T1  T2 
S  CP ln 
  CP ln
4 T1 T2
 T1 T2 
S  CP ln
T1  T2 
2
2
4 T1 T2
c. How does this expression show that this process is spontaneous?
The process will always be spontaneous if it is true that the argument to the log is greater
than 1, so that the entropy is positive. So we need to show that:
T  T 
1 1 2
2
4 T1 T2
Realize this form is totally symmetric in 1 and 2, so it no longer matters which is the hot one.
4 T1 T2  T1  T2   T12  2T1 T2  T2 2
2
0  T12  2T1 T2  T2 2  T1  T2 
So, yes it is true, because the difference of the two temperatures is squared, so the sign of the
difference does not matter, it will always be greater than 1.
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d. Does the hot system loose more entropy than the cold one gains, or vise versa?
The cold one (system1) must gain more entropy than the hot one loses:
T T 
1 1
   or    or T2  T1 
 T1 T2 
 T1 T2 
Hints: (1) Since the blocks are of the same material, they will have the same Cp. (2) Remember
that the system is isolated. (3) For part c, you can show this in various ways including
assuming numerical values for T1 and T2 in different cases.
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Chem 452B – HW 5 (B) Fall 2012
Q5) Steam is condensed at 100 °C and the water is cooled to 0 °C and frozen to ice. What is
the molar entropy change of the water? Consider that the average specific heat of liquid water
at constant pressure to be 4.2 J K-1 g-1, Hvap is 2258.1 J g-1 and Hmelt is 333.5 J g-1. (Ans: 154.6 J K-1 mole-1)
So follow the steps of condensing water, then moving from 100 to 0 C, and freezing the water.
Just add of the entropy changes for each step. As one goes from steam to water and water to
ice the process is both exothermic and therefore entropically disfavored (i.e. negative entropy
change):
S  S A  S B  SC
S 
H vap
TB
 CP ln
TM H melt  2258.1
273 333.5 


 4.2 ln

 18  154.5eu
TB
TM
373
273 
 373
Q6) Describe in one short phrase, with a short explanation, whether S will generally
increase or decrease in the following:
a. Neutralization of charges in an aqueous solution -- Specific entropy of the charges
themselves decrease because opposite charges are combining so the number of units is
dropping. Total entropy must increase of course. So solvent entropy increases.
b. A polar molecule is placed in a nonpolar solvent (or the opposite) -- entropy will go up,
slightly due to mixing, but solvent entropy can drop because solute puts constraints on
ways to arrange solvent. If the reaction is exothermic (which they usually are) the
environment could increase its entropy to offset system entropy decreases, as just
mentioned.
c. DNA mixed with cationic lipids in an aqueous solution (drawing a picture is
helpful) --- DNA is charged and surrounded with various counter ions rather than
mixing, overall the entropy must go up. But how it is partitioned is diffuclt to say,
presumably the organization of the lipid around the DNA decreases lipid entropy.
d. A phase change from a gas to a liquid --- increase in possible rearrangements and
exchanges of molecules, so entropy increases.
e. Increasing the number of molecules in the system (i.e. doubling the system) --- entropy
is extensive so the entropy is increased by having more of the same. The molar entropy
of a process does not change.
f. Increasing the temperature of the system --- Entropy always increases with increasing
Temperature, despite the definition of entropy.
g. Stretching a rubber band. --- You have to do work here, the system is not isolated so it
could go either way, but for a rubber band the entropy decreases because the strands of
the rubber band are more organized and have fewer opportunities to access various
configurations (so environment entropy must be increases).
Any spontaneous process in an isolated system happens because the total entropy of the system
goes up.
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Chem 452B – HW 5 (B) Fall 2012
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