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Unformatted text preview: ‐B
>
0
if
TB
>
TA)
and

 qB‐A
is
the
heat
lost
by
B
as
it
flows
into
A
(qB‐A
<
0
if
TB
>
TA).
If
we
denote
by
 MA
and
MB
the
molar
mass
of
A
and
B,
and
by
CPm(A)
and
CPm(B)
the
molar
 heat
capacities
at
constant
pressure
for
A
and
B,
respectively,
and
if
we
 note
that
the
heat
is
exchanged
at
constant
pressure,
then:
 qA‐B
=
ΔH(A)
=
(mA/MA)
CPm(A)
(TF
‐
TA)
and

 qB‐A
=
ΔH(B)
=
(mB/MB)
CPm(B)
(TF
‐
TB)
 This
allows
us
to
define
TF
by:
 mA m CPm (A) TA + B CPm (B) TB M MB TF = A 
 mA mB CPm (A) + CPm (B) MA MB Note
that,
if
A
and
B
are
made
of
the
same
material
and
have
the
same
 € mass,
then
the
final
temperature,
TF,
is
half
way
between
the
initial
 temperatures,
TA
and
TB.
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 103
 Now,
we
know
the
characteristics
of
the
final
state
(its
temperature)
and
 we
can
imagine
a
path
leading
from
the
initial
state
to
the
final
state,
which
 is
reversible.
A
reversible
path
could
be
a
constant
pressure
slow
heating
of
 A
from
TA
to
TF
and
a
constant
pressure
slow
cooling
of
B
from
TB
to
TF.
For
 such
processes,
the
changes
in
entropy
are
equal
to:
 T mA CPm (A) ln F MA TA 
 TF m ΔS(B) = B CPm (B) ln MB TB ΔS(A) = The
change
in
entropy
for
the
surroundings
is
zero
because
the
process
was
 € adiabatic
(total
heat
is
zero).
Therefore
the
total
change
in
entropy
for
 system
+
surroundings
is
equal
to
the
sum
of
the
entropy
change
for
A
and
 the
entropy
change
for
B.
 Let
us
take,
as
an
example
A
and
B
made
of
the
same
material
of
molar
 mass
100
g/mol.
Let
us
consider
the
block...
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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.

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