7B MT1 Sunday Review Session Solutions
Patrick Varilly and Steven Byrnes
1
Liphardt, Spring 05, Prob 2
a) Notice that
c
L
and
c
S
are plain heat capacities, not
speciﬁc
(per kg) heat capacities (look
at their units).
Deﬁne the following quantities.
Q
L
:= heat
gained
by liquid helium (negative if heat is
lost
by the helium);
T
L
:= instantaneous temperature of liquid helium;
T
L,i
:= initial temperature of the liquid helium;
T
L,f
:= ﬁnal temperature of the liquid helium;
S
L
:= entropy of the liquid helium
.
Also, deﬁne analogous quantities for the solid, with an
S
instead of an
L
subscript.
Now, since no work is being done, we have
d
Q
L
=
c
L
d
T
L
=
aT
3
L
d
T
L
.
Integrate this equation from the initial to the ﬁnal temperatures to get
Q
L
=
Z
T
L,f
T
L,i
aT
3
L
d
T
L
=
a
4
(
T
4
L,f

T
4
L,i
)
<
0
.
Numerically,
Q
L
=

0
.
024 J
.
b) Notice that
T
L,f
= 0
.
5 K now,
not
0
.
7 K. The calculation here is analogous to (b):
Q
S
=
Z
T
S,f
T
S,i
bT

2
S
d
T
S
=

b
(
T

1
S,f

T

1
S,i
)
.
Since the system is isolated, there is no net heat ﬂow:
Q
L
+
Q
S
= 0
.
Substituting our previous results in this last equation yields
a
4
(
T
4
L,f

T
4
L,i
)

b
(
T

1
S,f

T

1
S,i
)
= 0
.
When the liquid helium and the solid reach equilibrium, their temperatures will be equal.
We thus set
T
L,f
=
T
S,f
=
T
f
. Also, set
T
S,i
=
T
0
. We obtain the following equation
a
4
b
(
T
4
f

T
4
L,i
)
=
(
T

1
f

T

1
0
)
.
1