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23
21
avg
33
(1.38
10
J/K)[(32.0+273.15)K] = 6.32
10
J.
22
Kk
T
−−
==
×
×
The ratio
ε
/
K
avg
is (6.76
×
10
–20
J)/(6.32
×
10
–21
J) = 10.7.
27. (a) We use
ε
=
L
V
/
N
, where
L
V
is the heat of vaporization and
N
is the number of
molecules per gram. The molar mass of atomic hydrogen is 1 g/mol and the molar mass
of atomic oxygen is 16 g/mol so the molar mass of H
2
O is (1.0 + 1.0 + 16) = 18 g/mol.
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Mass, Heat

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