12. In each cycle
of its operation, a thermal engine absorbs 1000 J of heat from a large
heat reservoir at 400 K and discharges heat to another large heat sink
at 300 K. Calculate:
(a) The thermodynamic efficiency
of the heat engine, operated reversibly.
To determine the efficiency of a carnot engine
that is running reversibly we use the equation:
E = Th - Tl
Th
= 400 K - 300 K
400 K
= 1/4 = 25%
(b) The quantity of heat discharged
to the low-temperature sink each cycle.
In the engine the change in energy is equal to zero. Therefore we
use the equation:
dE = 0 = qab + qcd + w
qcd = qab + w
= 1000 J + (-1000J / 4)
= 750 J
(c) The maximum amount of work
the engine can perform each cycle.
The equation we use for work would be the product of the negative efficiency
and the energy.
W = -Eqab
= (-.25)(1000 J) = -250 J
26. All of the halogens react
directly with H2(g) to give binary compounds. The reactions are
F2(g) + H2(g) ---> 2 HF(g)
Cl2(g) + H2(g) ---> 2 HCl(g)
Br2(g) + H2(g) ---> 2 HBr(g)
I2(g) + H2(g) ---> 2 HI(g)
Using the data from Appendix D,
compute the change in entropy of formation of each reaction and identify
a periodic trend, if any.
The change in entropy for each of the reactants and products can be found
in the book. Using those for each reaction, we can determine the
change of entropy of formation by subtracting the reactants from the products.
Therefore the changes in entropy for each of the reactions are:
Reaction
Change in Entropy (J/K)
1
14.1
2
20.0
3
24.26
4
21.81
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