Water has potential energy. This can be expressed as the water mass times the elevation of its center of mass above some datum. The common event of water is to "flows downhill." (Some say the water "equilibrates with the Earth). So what becomes of the potential energy? Is energy conserved?
Water Seeks its Level The Waters of Earth exist predomiantly as liquid. Wherever water is found it resides with its surface flat and tangential to Earth (horizontal). Furthermore the surface is as physically low, or close to Earth as possible - subject to constraints that contain it.
In this example we consider an amount of water initially constrained within a tank. The "event" is that initial constraint is remoned. Water... freed from constraints moves toward Earth. Should such water encounter no constrants to its flow... that water becomes part of the sea.
The sketch depicts two steel tanks connected by a pipe with a valve (the event catalyst) that is closed, the tanks are open at their tops. Tank A is completely full of water and tank B is empty. What Second Depth in each tanks will result when the valve is opened and left open.
♦ The numerical answer to this physical event can be obtained by most students without using a plan. Numbers and answers, of course, are of less value than understanding. Physical situations of this text (like these tanks) are presented for the purpose of teaching the plan and method by which answers are obtained. This discussion (and those that follow) will show how the definitions, ideas and methods of thermodynamics constitute a plan.
Select an appropriate system and boundary: As system, we select all of the water. The system boundary is the outer surface of the water (its surface of contact with the air above and elsewhere where it contacts the steel of the constraining tank).
Describe and Sketch the Event: The event is an incremental change of the water from an initial equilibrium state (1) to a final equilibrium state (2). The event is initiated (t = 0+) when the valve is opened and water commences to flow from tank A to tank B. The event is defined to end with water in both tanks to depths, HA,2 and HB,2 (measured from the respective tank bottoms which are level). A sketch might be drawn identifying the system configurations (initial and final). The sketch we have will suffice.
What are the physical constraints? Initially the water is in gravitational equilibrium with Earth, subject to the structural constraint of the steel of tank A and the closed valve. In the final state, water is again in gravitational equilibrium with Earth subject to constraints of the tanks connected.
What does the mass equation say? Thermodynamics describes the mass of this event using a notation from calculus:
Initially the water occupies only tank A and tank B is empty. Finally there is water in both tanks. The mass equation in difference form is:
The water masses can be factored as density times volume: m = ρV.
The density of water is uniform, premitting the equation to be divided by density. Next we write the tank volumes in accord with their geometries and water depths: A being square, and B, cylindrical.
We run a check. Our steps and calculations above are correct. It is tempting to set HA,2 = HB,2 in the above equation. We know that is the case.
It is Earth, water (and the valve) that decide "what happens." For thousands of years water has been observed to "seeks its level." We argue that Earth's gravity causes contiguous water (released from constraints) to attain one surface. Setting the above depths equal, the final depth of the water (above tank bottoms) is:
H2 = 1.12 m.
What change of potential energy occurred? Potential energy of a body (or point mass) equals its mass times local gravity times its elevation relative to a selected datum for zero potential energy. Our selected datum is shown in the sketch, Z* = 0 . The water is not a body; it is a uniformly distributed fluid mass. Its potential energy can be expressed in terms of its center of mass (c.m.) can be used. Being uniformly distributed, its center of mass and its "center in space," have the same elevation.
PEwater,1 = mgo(Zc.m.,1 - Z*) = mgo(1 - 0)m
Potential energy is additive. After the event, the potential energies become:
PEwater,2 = mgo[(Z2,c.m. - Z*). Thus,
ΔPEwater = PEwater,2 - PEwater,1 = mgo[(Zc.m.,2 - Z*) - (Zc.m.,1) - Z*)] or
ΔPEwater = PEwater,2 - PEwater,1 = mgo[Zc.m.,2 - Zc.m.,1]
This equation shows that the datum taken of potential energy equal to zero (Z* = 0) is irrelevant. Next the tank geometries are regular. Initially the center of mass of the water is one meter above the bottom of tank A (or half of 2 meters). Later, (as calculated above) the water depth above the bottom in both tanks is 1.12 m. Thus, the second elevation of the water center of mass is: Z = (1.12/2)m or 0.56 m.
ΔPEwater = ρVgo(0.56m - 1.0m) c.m.,water
ΔPEwater = 1000 kg/m3 (8 m3)(9.81 m/s2)(0.58 m - 1.0 m) c. m., water
ΔPEwater = - 34,531 J
What became of this -34,531 Joules? Since mass moved above earth from a higher elevation (and PE) to a lower elevation (and Pe) we know ΔPE (i.e., PE2 - PE 1) must be negative. (Had the number been positive, we would know something was wrong). During the event, as water flowed, friction (mechanical dissipation) occurred. The friction caused the internal energy of the water to increase, evidenced by a slight increase of water temperature. Basic physics permits us to calculate the maximum possible temperature increase. The energy of the water is initially potential energy. The water retains that energy as it "seeks its level." The energy form changes from potential to internal No energy is "lost," the energy equation is:
ΔU + ΔPEwater = 0 or
mc(T2 - T1)water - 34,531 J = 0
The specific heat of water is: cwater = 4.2 kJ/(kg °C).
1000 kg/m3(8 m3)[4.2 kJ/(kg °C)](T2 - T1) - 34,531 J = 0.
Which solves as:
(T2 - T1)water ~ 1 °C
The initial temperature of the water was the same as the surrounding atmosphere. We assume it to be 25°C. We determined the second temperature of the water to be a maximum of one degree greater. With time, since temperatures equilibrate, and this thermal energy of the water distributes by the mechanism of heat to the surroundings until the temperature of the water in the tanks returns to 25°C. In conclusion, when water "seeks its level," the temperature of the universe increases (very very slightly).
If parts of the above solution seemed pedantic, it might be because you knew the "answer" to begin with. Solution of more difficult problems requires a diligent method which, at best, yields an approximate solution or, that failing, the diligent method points specifically to what "more information" is needed for solution. Although examples and illustrations of this writing are contrived, the underlying methods of solution and the understanding therein are genuine. Numerical answers provide a concreteness to our approximated solutions.