3.012 PS 2
1 of 6
10/4/05
3.012 PS 2
3.012
Issued: 09.20.05
Fall 05
Due:
09.30.05
THERMODYNAMICS
1. A sheet of manganese (2 moles) at room temperature (298 K) is placed in thermal contact
with a heat supply that slowly transfers 100,698 J of heat into the sample at constant
pressure. Use the following thermodynamic data for Mn to answer the questions below:
Mn has four solid phases,
α
,
β
,
γ
, and
δ
:
C
p
"
=
21.6
+
15.9
#
10
$
3
T
J
K
%
mole
T
trans
#
$
=
993
K
"
H
trans
#
$
%
=
2,010
J
mole
C
p
=
34.9
+
2.8
#
10
$
3
T
J
K
%
mole
T
trans
#
=
1373
K
"
H
trans
$
=
2,300
J
mole
C
p
=
44.8
J
K
#
mole
T
trans
#
=
1409
K
"
H
trans
$
=
1,800
J
mole
C
p
=
47.3
J
K
#
mole
a. Calculate the final temperature of the sample.
b. Calculate the total enthalpy change for this process.
c. Calculate the total entropy change for this process.
d. What phase (or phases) are present at equilibrium at the end of this process?
a. Calculation of the final temperature is made by using the heat capacity data to determine how
much heat is required to heat up and through the several phase transitions present. The
calculation is readily carried out by simply determining how much heat is needed for heating
to each phase transition:
Starting out at 298 K, we first heat the sample (
α
phase at room temperature) to the first phase
transition
α
>
β
, at 993K:
q
298
K
"
993
K
=
n
C
p
dT
298
993
#
=
(2 moles)(21.6
+
15.9
$
10
%
3
T
J
mole
&
K
)
dT
298
993
#
=
44,290
J
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10/4/05
…clearly, we have more heat to use, so we continue. The phase transition consumes a small
amount of heat:
q
298
K
"
993
K
=
n
#
H
$
"
%
=
(2)(2,010)
=
4,020
J
We continue summing up the heat for the next few transitions in a similar manner:
q
993
K
"
1373
K
=
n
C
p
#
dT
993
1373
$
=
(2 moles)(34.9
+
0.0028
T
J
mole
%
K
)
dT
993
1373
$
=
29,041
J
q
"
#
=
n
%
H
#
=
(2)(2,300)
=
4,600
J
Our running total for heat transferred to reach the gamma phase ast 1373K is now 81,951 J. This
leaves 18,747 J of heat yet to be consumed. Continuing on to the next transition:
q
1373
K
"
2300
K
=
n
C
p
dT
1373
1409
$
=
(2 moles)(44.8
J
mole
%
K
)
dT
1373
1409
$
=
3,226
J
q
#
=
n
%
H
#
=
(2)(1,800)
=
3,600
J
We have 11,701 J of heat yet to use up. We determine the final temperature by integrating to use up
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 Fall '13
 Entropy, Heat, Heat Transfer, Magnetic Field, NC

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