Chem 6A 2011 (Sailor)
QUIZ #5
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VERSION A KEY
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Section Number:
Some useful constants and relationships:
Specific heat capacities (in J/g
.
K): H
2
O
(l)
= 4.184; Al
(s)
= 0.900; Cu
(s)
= 0.387; Steel
(s)
= 0.45
101.325 J = 1 L
.
atm
1 atm = 760 Torr
1J = 1kg
.
m
2
/s
2
1 eV = 1.6022 x 10
19
J
R = Ideal gas constant: 0.08206 L
.
atm
.
mol
1 .
K
1
= 8.31451 J
.
mol
1 .
K
1
Avogadro constant:
6.022 x 10
23
mole
1
Planck's constant = h = 6.6261 x 10
34
J
.
s
c = speed of light: 3.00 x 10
8
m/s
R
H
= 1.097 x 10
2
nm
1
C
2
= second radiation constant = 1.44 x 10
2
K
.
m
q = C
p
m
Δ
T
T
λ
max
=
1
5
C
2
Emitted power (W)
Surface area (m
2
)
=
(
cons
tan
t
)
T
4
e
=
mc
2
c
=
λν
1
=
R
H
1
n
1
2
−
1
n
2
2
Λ
Ν
Μ
Ξ
Π
Ο
E
=
h
ν
E
=
hc
E
(in Joules)
=
−
2.18
×
10
−
18
Z
2
n
2
Λ
Ν
Μ
Ξ
Π
Ο
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QUIZ #5
1.
(a) Calculate the heat (q) when 22.0 g of solid water ice cools from 11.0 °C to
97.0 °C.
Set up but do not solve; circle your answer.
The following table of
thermochemical data for H
2
O may be useful. (20 pts)
this was problem 6.23 in the
text
Heat of fusion
6 kJ/mol
Heat of vaporization
48
kJ/mol
heat capacity of
H
2
O
(s)
2.087 J/g
.
K
heat capacity
of H
2
O
(l)
4.184 J/g
.
K
heat capacity of H
2
O
(g)
2.080
J/g
.
K
standard heat of
formation of H
2
O
(l)
241.8
kJ/mol
q
=
C
p
m
Δ
T
=
2.087
J
g
⋅
K
×
22.0
g
×
(
−
97.0
+
11)
Numerically this works out to 3,949 J.
15 pts for correct setup. OK if they get the sign wrong.
2 pts if they use the wrong value of heat capacity (4.184 is for liquid water, 2.080 is
for gas phase water)
(b) Is this process exothermic or endothermic (circle one)?
exothermic 5 pts all or
nothing
2.
Calculate the final temperature when a steel bolt, initially at 100.0 °C and weighing
50.0 g, is placed in 37 g of liquid water, initially at 11.0 °C.
The relevant
thermochemical data are given on the cover page of this quiz. (10 pts)
this was a
variation of problems 6.26 and 6.28 in the text
heat lost by bolt =  (heat gained by water), T
final
is the same for both
applying the expression
q
=
C
p
m
Δ
T
to both,
C
bolt
m
bolt
Δ
T
=
−
C
water
m
water
Δ
T
C
bolt
m
bolt
(
T
final
−
T
bolt
)
=
−
C
water
m
water
(
T
final
−
T
water
)
multiplying through,
C
bolt
m
bolt
T
final
−
C
bolt
m
bolt
T
bolt
=
−
C
water
m
water
T
final
+
C
water
m
water
T
water
collect terms, simplify,
C
bolt
m
bolt
T
final
+
C
water
m
water
T
final
=
C
water
m
water
T
water
+
C
bolt
m
bolt
T
bolt
T
final
(
C
bolt
m
bolt
+
C
water
m
water
)
=
C
water
m
water
T
water
+
C
bolt
m
bolt
T
bolt
solve for T
final
T
final
=
C
water
m
water
T
water
+
C
bolt
m
bolt
T
bolt
C
bolt
m
bolt
+
C
water
m
water
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 Spring '08
 Pomeroy
 Thermodynamics, Enthalpy, Avogadro constant, Cbolt mbolt, Cbolt mbolt Tbolt

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