1. Heat Capacity
If a system experiences a change of temperature from Ti to Tf during the transfer of Q units of heat, the average (mean) heat capacity of the system is defined as the ratio:
The true heat capacity, C :
Unit: JK-1
Cannot be interpreted as the derivative of Q with respect to T
d’Q means “a small flow of heat” and dT is the corresponding change in temperature
internal energy can be changed either by heat or work.
Practically, easier to use heat than work. So when systematic experiments were performed to measure the capability of a substance to store internal energy, heat rather than work was used, and the results came to be known as the heat capacity of the sample. The term “heat capacity” implies that a substance can hold heat, which is completely false. Heat is not a function of the thermodynamic state of a system; internal energy is! The proper expression should be internal energy capacity.
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2. Two important cases for hydrostatic systems:
Heat capacity at constant volume:
Heat capacity at constant pressure:
depending on the process the system undergoes during the heat transfer.
V=constant M ΔT
CV=QV/ΔT Qv
P=constant
CP=QP/ΔT QP
(2)
(1)
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3. Each simple system has its own heat capacities.
Heat capacity
Symbol
Hydrostatic system
At constant pressure
At constant volume CP CV
Stretched wire
At constant tension
At constant length CF CL
Surface
At constant surface tension
At constant area Cσ CA
Electrochemical cell
At constant emf
At constant charge CE
Dielectric slab
At constant electric field
At constant polarisation
Paramagnetic rod
At constant magnetic field
At constant magnetisation CH CM
Each heat capacity is a function of two variables
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4. The concept of heat capacity applies to a given system.
The specific heat capacity, the heat capacity per unit mass or per unit mole, is the characteristic of the material of which the system is composed
Specific heat capacity
Molal specific heat capacity
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5. At low temperature, cp≈cv
Variation of molal specific heat capacities cp and cv with temperature for copper at constant pressure of 1 atm
At low temperature, cp≈cv
near absolute zero, both drop rapidly to zero
This behaviour is characteristic of most solids, although the temperature at which the sharp drop occurs varies widely
At high temperature,
cp continues to increase
cv  ≈25 x 103 J kilomole-1 K-1
cv of many solids approach this value at high temperatures
 the Dulong and Petit value
Note: R=8.31x103 J kilomole-1 K-1
Thus, at high T, cv for solids≈ 3R
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6. The change with pressure of cp and cv for mercury, at constant temperature.
The pressure variation is relatively much smaller than the variation with temperature
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7. Some values of cp and cv for gases expressed in terms of R
Monotomic gases:
cp/R ≈5/2=2.50
cv/R≈3/2=1.50
Diatomic gases:
cp/R ≈7/2=3.50
cv/R≈5/2=2.50
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8. Over a temperature interval in which C can be considered constant,
The total quantity of heat flowing into a system, in any process, is given by
Over a temperature interval in which C can be considered constant,
Larger C  smaller ΔT
Therefore, by making C very large, ΔT can be made as small as we please.
A system of very large C  a heat reservoir
heat flow into or out of the reservoir can be as large as we please, without any change in the temperature
Thus, any reversible process carried out by a system in contact with a heat reservoir is isothermal
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9. Heats of transformation.
In some situations, a substance (system) can absorbed heat without increasing its temperature at all.
This normally happens at a phase transformation (e.g. boiling water)
Technically, the heat capacity is then infinite:
(during a phase change)
However, you might still want to know how much heat is required to melt or boil the substance completely.
This amount, divided by the mass, is called the latent heat of transformation.
e.g. melting ice : l = 3.33 x 105 J kg-1
boiling water l = 22.6 x 105 J kg-1
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10. Phase changes are always associated with changes in volume
Work is always done on or by the system in a phase change
(except at the critical point where the specific volumes of liquid and vapour are equal)
If the change occurs at constant T and constant P, and no other work is done besides the usual constant pressure expansion or compression:
Specific work done by the system :
From 1st law:
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11. (u+Pv) occurs frequently in thermodynamics
Since u, P, v are all properties of the system, (u+Pv) also defines a property
So, define specific enthalpy :
Unit: J kg-1 or J kilomole-1
Solid to vapour (sublimation)
Solid to liquid (fusion)
Liquid to vapour (vapourisation)
Since enthalpy is a state function:
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12. Adding PV onto energy gives the quantity called the enthalpy, H H=U+PV
Constant pressure processes occur often (both naturally and in the lab.)
Keeping track of compression-expansion work can be a pain after a while!
A convenient trick:
Instead of always talking about energy content of the system, we can agree to always add in the work needed to make room for it (under constant pressure, usually 1 atm).
This work is PV, the pressure of the environment x total volume of the system (i.e. the space needed to clear out to make room for it)
Adding PV onto energy gives the quantity called the enthalpy, H
H=U+PV
This is the total energy needed to create the system out of nothing and put it into the environment.
Put in another way, to annhilate the system, the energy that could be extracted is not just U, but also the work (PV) done by the atmosphere as it collapse to fill the vacuum left behind.
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13. To create the rabbit out of nothing and place it on the table, the magician must summon up not the energy U of the rabbit, but also some additional energy, equal PV, to push the atmosphere out of the way to make some room. The total energy required is the enthalpy, H=U+PV.
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14. So the change in enthalpy during a constant-pressure process is:
Suppose, some change takes place in the system –(add heat, or chemicals react, or whatever,…)- while P=constant.
The energy, volume, and enthalpy can all change by amounts = ΔU, ΔV, and ΔH
The new enthalpy is:
H + ΔH = (U + ΔU) + P(V + ΔV)
=(U + PV) + (ΔU + PΔV)
= H + (ΔU + PΔV)
So the change in enthalpy during a constant-pressure process is:
ΔH = ΔU + PΔV
This says that the energy can increase for 2 reasons:
either (a) because the energy increases, or
(b) because the system expands and work is done on the atmosphere to make room for it.
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15. or: ΔU = Q – (PΔV) - Wother
Recall: 1st law: ΔU = Q – W
or: ΔU = Q – (PΔV) - Wother
Other type of work, e.g. electrical, chemical
(work done on system)
compression/expansion work
Therefore: ΔH = Q – Wother (constant P)
i.e. the change in enthalpy is caused only by heat and other forms of work, not by compression-expansion work.
In other words,: can forget about compression-expansion work if we deal with enthalpy instead of energy.
If no “other” types of work is done, the change in enthalpy tells us directly how much heat has been added to the system.
For simple case of raising an object’s temperature,the change in enthalpy per degree, at constant pressure is the same as the heat capacity at constant pressure, CP :
(“enthalpy capacity”)
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16.  ΔH when boiling 1 g of water is (40,660 J) / (18 g) = 2260 J.
Chemistry books are full of tables of ΔH for more dramatic processes: phase transformation, chemical reactions, ionisation, dissolution in solvents, etc…
Example (a)
From standard table: ΔH for boiling water for 1 mole at 1 atm is 40,660 J.
since, 1 mole H2O = 18 g
 ΔH when boiling 1 g of water is (40,660 J) / (18 g) = 2260 J.
However, not all of this energy ends up in vapourising water.
The volume of 1 mole of water vapour = RT/P (the initial volume is negligible).
therefore, work needed to push the atmosphere away is:
PV = RT = (8.31 J K-1)(373 K) = 3100 J
(this is only 8% of 40,660 J of energy put in)
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17. Example (b)
Chemical reaction: Hydrogen combine with oxygen gas to form liquid water:
H2 + ½O2  H2O
for each mole of water produced, ΔH = -286 kJ (in the tables this is known as the enthalpy of formation of water)
So if we simply “burn” 1 mole of hydrogen, then 286 kJ is the amount of heat we get out.
Nearly all of this energy comes from the thermal and chemical energy of the molecules themselves.
A small amount comes from the work done by the atmosphere as it collapses to fill the space left behind by the consumed gases.
Can some of the 286 kJ be extracted as work (perhaps electrical work) rather than heat?
In general the answer is that much of the energy from a chemical reaction can be extracted as work, but there are limits.
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