(b) There are two alternative explanations of the efficiency of a Carnot
cycle:
(i) Higher reversible work with lower T B is due to stronger entropy
growth potential, thus, a greater amount of heat extracted.
(ii) Or, higher reversible work with lower T B is due to smaller heat to
be discarded to the sink.
Although both are factually correct for the case at hand, explain why the
first of the two is a preferred explanation.
8:6 Imagine Rumford’s cannon experiment: by immersing a cannon barrel in a
water tank of 100 gallons, with water initially at 30 °C and the barrel was
bored with a specially blunted boring tool until water starts to boil. The
measured work input is 110,694 kJ. Determine the entropy growth at the
end of the experiment.
Now consider a long time afterward at the completion of the boring when
the water in the tank is cooled down to the surrounding temperature of 30 °
C. What will be the eventual entropy growth? Explain that the eventual
entropy growth is the result of the boring process and the cooling process in
water (determine the value of entropy growth due to the cooling process).
Going back to the initial setup of a “system” as the work reservoir of
110,694 kJ and a heat reservoir at 30 °C, we consider the process as a
spontaneous dissipation of mechanical energy of 110,694 kJ into heat.
Without the details of the actual processes described above, what can you
tell in terms of the entropy growth and the entropy growth potential of this
mechanical energy ! heat example?
8:7 Thermal spontaneity example: The above cooling part of Prob. 8.6 can be
analyzed in a similar manner to Prob. 8.2 and Prob. 8.3. Given a system of
100 gallons of water at 100 °C in the surroundings (heat reservoir) at 30 °C.
What are the entropy growth and the entropy growth potential.
Spontaneous event: What is the entropy growth in a spontaneous cooling
event?
Reversible event (Heat ! mechanical energy example): Conversion, more
precisely the extraction, of heat into mechanical energy is possible only
with the existence of thermal spontaneity under the control of reversible-like
management. Describe such a reversible event and explain why such an
event entails no entropy growth.
Explain why both events are driven by entropy growth potential.
8:8 Show in detail steps the derivation of the generalized Carnot–Kelvin
formula.
8:9 An isolated composite system consists of gaseous subsystem
(1) and subsystem
(2) (same gas in both) of equal mole numbers (N
(1) = N
(2) = N),
initially at p and T 1 , and p and T 2, respectively. They are insulated from
each other but connected pneumatically with a valve in closed position
8.8 Entropy Growth Potential and Reversibility’s Triadic Framework
229
cycle:
(i) Higher reversible work with lower T B is due to stronger entropy
growth potential, thus, a greater amount of heat extracted.
(ii) Or, higher reversible work with lower T B is due to smaller heat to
be discarded to the sink.
Although both are factually correct for the case at hand, explain why the
first of the two is a preferred explanation.
8:6 Imagine Rumford’s cannon experiment: by immersing a cannon barrel in a
water tank of 100 gallons, with water initially at 30 °C and the barrel was
bored with a specially blunted boring tool until water starts to boil. The
measured work input is 110,694 kJ. Determine the entropy growth at the
end of the experiment.
Now consider a long time afterward at the completion of the boring when
the water in the tank is cooled down to the surrounding temperature of 30 °
C. What will be the eventual entropy growth? Explain that the eventual
entropy growth is the result of the boring process and the cooling process in
water (determine the value of entropy growth due to the cooling process).
Going back to the initial setup of a “system” as the work reservoir of
110,694 kJ and a heat reservoir at 30 °C, we consider the process as a
spontaneous dissipation of mechanical energy of 110,694 kJ into heat.
Without the details of the actual processes described above, what can you
tell in terms of the entropy growth and the entropy growth potential of this
mechanical energy ! heat example?
8:7 Thermal spontaneity example: The above cooling part of Prob. 8.6 can be
analyzed in a similar manner to Prob. 8.2 and Prob. 8.3. Given a system of
100 gallons of water at 100 °C in the surroundings (heat reservoir) at 30 °C.
What are the entropy growth and the entropy growth potential.
Spontaneous event: What is the entropy growth in a spontaneous cooling
event?
Reversible event (Heat ! mechanical energy example): Conversion, more
precisely the extraction, of heat into mechanical energy is possible only
with the existence of thermal spontaneity under the control of reversible-like
management. Describe such a reversible event and explain why such an
event entails no entropy growth.
Explain why both events are driven by entropy growth potential.
8:8 Show in detail steps the derivation of the generalized Carnot–Kelvin
formula.
8:9 An isolated composite system consists of gaseous subsystem
(1) and subsystem
(2) (same gas in both) of equal mole numbers (N
(1) = N
(2) = N),
initially at p and T 1 , and p and T 2, respectively. They are insulated from
each other but connected pneumatically with a valve in closed position
8.8 Entropy Growth Potential and Reversibility’s Triadic Framework
229