Ifweassumethatattemperaturetthe

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s
to
have
the
same
mass
(1000
g)
 and
let
us
consider
the
molar
heat
capacity
to
be
24.44
J.mol‐1.K‐1.

If
the
 initial
temperature
of
the
blocks
are
respectively,
0°C
and
100°C,
then
the
 final
temperature,
TF
=
50°C
(see
above).
The
entropy
change
for
the
two
 blocks
will
be:
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 104
 T m T mA CPm (A) ln F + B CPm (B) ln F MA TA MB TB 
 323 1000 323 −1 ΔS = (24.44)ln + ln = 5.93 J.K 100 373 273 ΔS = ΔS(A) + ΔS(B) = Note
that
since
ΔSSurroundings
=
0,
then
ΔSUniverse
=
ΔS.
Therefore
ΔSUniverse
is
 € positive,
in
accord
with
the
requirement
that
ΔSUniverse

>
0
for
a
 spontaneous
process.
Indeed,
heat
transfer
and
temperature
equilibration
 processes
are
spontaneous.
 
 Example
#2:
 Consider
what
happens
when
you
mix
under
a
pressure
of
1
bar,
500
g
of
 ice
at
‐10°C
with
100
g
of
H2O
vapor
at
+110°C.

Assume
the
mixing
takes
 place
in
a
thermally
insulated
container,
so
that
the
heat
transfer
occurs
 exclusively
between
the
ice
and
the
vapor.
What
is
the
final
state
of
the
 system
and
what
is
the
change
in
entropy
for
the
process
(system
+
 surroundings)?

 Here,
we
will
assume
that
the
final
state
of
the
system
is
liquid
water
at
 some
temperature
TF
(between
0
and
100oC).
The
latent
heat
of
 vaporization
(standard
enthalpy
change
for
vaporization)
is
equal
to

 44
kJ/mol
and
that
for
the
fusion
is
6
kJ/mol.
These
processes
occur
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 105
 reversibly
at
1
bar
at
373
K
and
273
K,
respectively.

The
molar
heat
 capacity
at
constant
pressure
of
wat...
View Full Document

This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

Ask a homework question - tutors are online