# Zeroth Law of Thermodynamics

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27 Οκτ 2013 (πριν από 4 χρόνια και 8 μήνες)

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Zeroth Law of
Thermodynamics
and the concept of temperature and “heat flow”
Zeroth Law of Thermodynamics
There is a state function, temperature -
T
, which has the
following property:
Zeroth Law of Thermodynamics
There is a state function, temperature -
T
, which has the
following property:
!
If heat is added to a closed system,
T
will increase
.
Conversely, if heat is extracted from a closed system,
T
will decrease.
Zeroth Law of Thermodynamics
There is a state function, temperature -
T
, which has the
following property:
T
1
>
T
2
!
If two closed systems with different temperatures are
brought into thermal contact, the heat will “flow” from the
system of higher
T
to the system of lower
T
until the two
systems reach the same intermediate temperature*.
!
If heat is added to a closed system,
T
will increase
.
Conversely, if heat is extracted from a closed system,
T
will decrease.
T
2
T
1
* Together, the two
systems are an isolated
system.
Zeroth Law of Thermodynamics
There is a state function, temperature -
T
, which has the
following property:
T
final
T
1
>
T
final

>
T
2
T
final
!
If two closed systems with different temperatures are
brought into thermal contact, the heat will “flow” from the
system of higher
T
to the system of lower
T
until the two
systems reach the same intermediate temperature*.
!
If heat is added to a closed system,
T
will increase
.
Conversely, if heat is extracted from a closed system,
T
will decrease.
* Together, the two
systems are an isolated
system.
Zeroth Law of Thermodynamics
Another way of stating this is:
Two closed systems are said to be in a state of equilibrium if
when they are brought in thermal contact there is no net
change in their state functions.
T
final
T
1
>
T
final

>
T
2
T
final
* Together, the two
systems are an isolated
system.
Heat Capacity
The zeroth law does not imply that the intermediate
temperature will be an average of the two temperatures.
Actually, it doesn’t even imply a “linear” temperature scale,
only a monotonic one. Thus, a scale must be established by
other means.*
T
final
T
final
T
1
>
T
final

>
T
2
*This is established with
the 3rd Law.
Heat Capacity
For example, assume system 1 has twice the mass of
system 2 and the two system are composed of identical
materials. Then one might expect the change in temperature
for system 2 to be twice as great as for system 1.
(With the proviso the starting temperature were fairly
close.)
T
final
T
final
T
1
!
T
final
.
T
final
!T
2

Heat Capacity
For example, assume system 1 has twice the mass of
system 2 and the two system are composed of identical
materials. Then one might expect the change in temperature
for system 2 to be twice as great as for system 1.
But what if they are composed of different mass
and
different
materials?
T
final
T
final
T
1
!
T
final
.
T
final
!T
2

Heat Capacity
For example, assume system 1 has twice the mass of
system 2 and the two system are composed of identical
materials. Then one might expect the change in temperature
for system 2 to be twice as great as for system 1.
But what if they are composed of different mass
and
different
materials?
This requires the use of the concepts of heat transfer or
“heat flow” and heat capacity.
T
final
T
final
T
1
!
T
final
.
T
final
!T
2

Heat Capacity
Definition: Heat transfer or “heat flow”,
q
, is the amount of
energy transferred from one closed system to another when
brought into thermal contact.
T
final
T
final
T
1
<
T
2

Heat Capacity
Definition: Heat transfer or “heat flow”,
q
, is the amount of
energy transferred from one closed system to another when
brought into thermal contact.
Definition: Heat Capacity,
C
, is defined by the equation:
q
=
C
Ä
T
T
final
T
final
T
1
<
T
final
<
T
2

Heat Capacity
Definition: Heat transfer or “heat flow”,
q
, is the amount of
energy transferred from one closed system to another when
brought into thermal contact.
Definition: Heat Capacity,
C
, is defined by the equation:
q
=
C
Ä
T
If the system consist of a pure substance, chances are that
the heat capacity has been measured and tabulated. It is
tabulated, however, on the basis of heat capacity per gram or
heat capacity per mole of material. These are given the
symbols and names:
S=
heat capacity per gram called “specific heat”
C
= heat capacity per mole called “molar heat capacity”
Heat Capacity
A summary to this point: Definition: Heat transfer or “heat
flow”,
q
, is the amount of energy transferred from one closed
system to another when brought into thermal contact.
Definition: Heat Capacity,
C
:
q
=
C
Ä
T
S=
heat capacity per gram called “specific heat”
C
= heat capacity per mole called “molar heat capacity”
Heat Capacity
A summary to this point: Definition: Heat transfer or “heat
flow”,
q
, is the amount of energy transferred from one closed
system to another when brought into thermal contact.
Definition: Heat Capacity,
C
:
q
=
C
Ä
T
S=
heat capacity per gram called “specific heat”
C
= heat capacity per mole called “molar heat capacity”
C
, is given a subscript “p” of “v”
depending upon whether the conditions are constant
pressure or constant volume.
C
p
=
molar heat capacity at constant pressure.
C
v
=
molar heat capacity at constant volume.
In chemistry
C
p
is used most often.
Heat Capacity
Example 1: Calculate the amount of heat required to increase
the temperature of a 35.0 g piece of lead from 0.0EC to 30.0EC.
The molar heat capacity for lead:
C
p

= 26.1 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of lead:
n
= (35.0 g)/(207.2 g mol
–1
)
n
= 0.169 mol
Example 1: Calculate the amount of heat required to increase
the temperature of a 35.0 g piece of lead from 0.0EC to 30.0EC.
The molar heat capacity for lead:
C
p

= 26.1 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of lead:
n
= (35.0 g)/(207.2 g mol
–1
)
n
= 0.169 mol
2) Calculate the total heat capacity:
C
=
n
C
p
C
=
(0.169 mol)(26.1 J mol
–1
K
–1
)
C
= 4.42 J K
–1
Example 1: Calculate the amount of heat required to increase
the temperature of a 35.0 g piece of lead from 0.0EC to 30.0EC.
The molar heat capacity for lead:
C
p

= 26.1 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of lead:
n
= (35.0 g)/(207.2 g mol
–1
)
n
= 0.169 mol
Example 1: Calculate the amount of heat required to increase
the temperature of a 35.0 g piece of lead from 0.0EC to 30.0EC.
The molar heat capacity for lead:
C
p

= 26.1 J mol
–1
K
–1
3) Substitute into
q = C
Ä
T
:
q =
(4.42 J K
–1
)(30.0 - 0.0)K
q =
133 J
2) Calculate the total heat capacity:
C
=
n
C
p
C
=
(0.169 mol)(26.1 J mol
–1
K
–1
)
C
= 4.42 J K
–1
Heat Capacity
Example 1: Calculate the temperature increase when 750 J of
heat is applied to 150 g of NaCl.
The molar heat capacity for NaCl:
C
p

= 49.9 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of NaCl:
n
= (150 g)/(58.5 g mol
–1
)
n
= 2.56 mol
Example 1: Calculate the temperature increase when 750 J of
heat is applied to 150 g of NaCl.
The molar heat capacity for NaCl:
C
p

= 49.9 J mol
–1
K
–1
Heat Capacity
1) Calculate the number of moles of NaCl:
n
= (150 g)/(58.5 g mol
–1
)
n
= 2.56 mol
Example 1: Calculate the temperature increase when 750 J of
heat is applied to 150 g of NaCl.
The molar heat capacity for NaCl:
C
p

= 49.9 J mol
–1
K
–1
2) Calculate the total heat capacity:
C
=
n
C
p
C
=
(2.56 mol)(49.9 J mol
–1
K
–1
)
C
= 128 J K
–1
Heat Capacity
1) Calculate the number of moles of NaCl:
n
= (150 g)/(58.5 g mol
–1
)
n
= 2.56 mol
Example 1: Calculate the temperature increase when 750 J of
heat is applied to 150 g of NaCl.
The molar heat capacity for NaCl:
C
p

= 49.9 J mol
–1
K
–1
3) Substitute into
q = C
Ä
T
:
750 J
=
(128 J K
–1

T
ÄT =
5.86 K (or EC
)
2) Calculate the total heat capacity:
C
=
n
C
p
C
=
(2.56 mol)(49.9 J mol
–1
K
–1
)
C
= 128 J K
–1
Zeroth Law of
Thermodynamics
and the concept of temperature and “heat flow”
THE END