McKay, Donald. "Front matter" Multimedia Environmental Models ...

heat capacity of 0.38 J/g°C requires absorption of only 0.38 J to cause the same rise
in temperature. Hydrogen gas has a large heat capacity of 14.3 J/g°C and thus
requires a great deal of heat to raise its temperature. These substances differ markedly
in their temperature response when heat is added. If 1000 J are added to equal masses
of 1 g of these substances, the copper becomes much hotter by 263°C (or 100/0.38),
while the water only heats up by 24°C (or 100/4.2) and the hydrogen by only 7°C
(or 100/14.3). Hydrogen and water can thus absorb or “soak up” larger quantities
of heat without becoming much hotter.
The fugacity capacity is similar. Phases of high Z (possibly sediments or fish)
are able to absorb a greater quantity of solute yet retain a low fugacity. It follows
that solutes will tend to partition into these high Z phases and build up a substantial
concentration yet retain a relatively low fugacity. Conversely, phases with low Z
values will tend to experience a large increase in f following absorption of a small
quantity of solute. A substance such as DDT is readily absorbed by fish and achieves
a high concentration at low fugacity. The Z value of DDT in fish is large. On the
other hand, DDT is not readily absorbed by water; indeed, it is hydrophobic or
“water hating.” Its Z value in water is very low.
This analogy between heat and fugacity capacity is perhaps best illustrated by
the following pair of numerically identical examples, the fugacity quantities being
given in parentheses.
Worked Example 5.7
A system consists of three phases 10 g of water (10 m 3 of water) of heat capacity
4.2 J/g°C (fugacity capacity 4.2 mol/m 3 Pa), 5g of copper (5 m 3 of air) of heat
capacity 0.38 J/g°C (fugacity capacity 0.38 mol/m 3 Pa), and 1 g of hydrogen (1 m 3
of sediment) of heat capacity 14.3 J/g°C (i.e., fugacity capacity mol/m 3 Pa). To this
system is added 582 J of heat (582 mol of solute). What is the heat (solute)
distribution at equilibrium, and what is the rise in temperature (fugacity) and heat
concentrations in J/g (concentrations in mol/m 3 ). We assume for simplicity that the
initial temperature is 0°C, and the initial concentrations are also zero. (Note that Z
for a solute in air never has the above value.)
When approaching equilibrium, the temperatures (fugacities) will rise equally
to a new common value at T °C, (f) such that the amount of heat (solute) in each
phase will be
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mass (g) ¥ heat capacity ¥ T or [volume (m 3 ) ¥ Zf]
Thus, the total will be the summation over the three phases, i.e.,
Thus,
582 = 10 ¥ 4.2 ¥ T + 5 ¥ 0.38 ¥ T + 1 ¥ 14.3 ¥ T
T = 582/(10 ¥ 4.2 + 5 ¥ 0.38 + 1 ¥ 14.3) = 10°C (Pa)