PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Class Activity - Class 7 February 7, 2006
Name___________ANSWERS________________
Do problem 7-3 from the textbook. Since many of you do not have the textbook with you, it is
given here. (Hint: Do not believe all the answers in the back of the book.)
Calculate the change in the entropy of the universe as a result of each of the following processes:
(a) A copper block of mass 0.4 kg and heat capacity 150 J/K at 100ºC is placed in a lake at 10ºC.
(b) The same block at 10ºC is dropped from a height of 100 m into the lake.
(c) Two similar blocks at 100ºC and 10ºC are joined together. (Hint: See problem 7-8, or today’s
lecture notes.
(d) One kilomole of a gas at 0ºC is expanded reversibly and isothermally to twice its initial
volume.
(e) One kilomole of a gas at 0ºC is expanded reversibly and adiabatically to twice its initial
volume.
T 
dT
 283 K 
 CP ln  2   (150 J/K) ln 
  41.4 J/K
T T
 373 K 
 T1 
Q CP (T1  T2 ) (150 J/K)(373 K  283 K )
Slake 


 47.7 J/K
T2
T2
283 K
Suniverse = Sblock + Slake = – 41.4 J/K + 47.7 J/K
(a) Sblock  CP 
T2
1
(b) Sblock = 0
Q mgh (0.4 kg) (9.8 m/s 2 )(100 m)
Slake 


 1.39 J/K
T2
T2
283 K
Suniverse = Sblock + Slake = 0 + 1.39 J/K
Suniverse = 6.3 J/K
Suniverse = 1.39 J/K
 T T 
 373 K  283 K 
  2.85 J/K
(c) Sblocks  2CP ln  1 2   2(150 J/K)ln 

 2 TT 
1 2 
 2 (373 K)(283 K) 

Ssurroundings = 0
Suniverse = Sblocks + Ssurroundings = 2.85 J/K + 0
Suniverse = 2.85 J/K
(d) Since the process is reversible,
Suniverse = 0
For the isothermal process, dU = 0, so đQ = đW = PdV.
2V0 PdV
2V0 RT dV
2V0 dV
 2V 
S gas  

 R
 R ln  0   R ln 2  5.76  103 J/K
V0
V0
V0
T
V T
V
 V0 
Ssurroundings = – 5.76 ×103 J/K
(e) Since the process is reversible,
Sgas = 0
and
Ssurroundings = 0
Suniverse = 0