Energy not conserved?

Here is a nifty problem with a seeming violation of the conservation of energy: The sum of the 4 E’s should be zero, because we go around a complete cycle and end up exactly the same as at the beginning. Of course ignore all lossy things like friction and heat transfer to or from the surroundings. The guy that cooked up this problem hopes to show that energy is not conserved.

1) Raise a limp balloon, the ideal insulator, containing liquid water at the boiling point. Let the volume be negligible. E₁ = mgh.

2) Boil the water, E₂ = L_V m. Now the volume is increased to V, so there is a significant buoyant force F_B on it.

3) Pull the balloon down with force F_B – mg. E₃ = (F_B- mg)h.

4) Condense the water taking away energy: E₄ = -L_V m

What is going on? These do not add up to zero! The sum is F_Bh.

Solution

When you boil water, part of the energy goes to breaking the inter-molecular bonds, and part to expanding against pressure P of the atmosphere. So we should change the expression in 2) to E₂ = Lm + PV (L is somewhat smaller than the usual L_V), and the expression in 4) to E₄ = -Lm -(P+DP)V, where DP is the change of pressure while moving down distance h.

Remember that buoyant force depends on the difference in P between the top and bottom of an object. A rectangular solid with horizontal area A and height y has F_B equal to the difference in pressures DP’ times A. But pressure changes in proportion to distance if the density of the fluid is constant (and if it is not constant this can be fixed up with calculus) so DP/h = DP’/y, so DP’ = yDP/h, and F_B = DP’A = (yDP/h)A. But yA = V, so F_B h = DPV. Now the sum of the energy changes = 0. ( Note: F_B is positive, so DP is P_o-P in this analysis.)

1) mechanical work E₁ = mgh

2) heat added E₂ = Lm + PV

3) mechanical work E₃ = (F_B - mg)h = DPV – mgh = (P_o – P)V - mgh

4) heat taken away E₄ = -Lm - P_oV

Now the sum of E₁ through E₄ = 0.

Please note, however, that this is a fluke that it came out exactly zero, because L is slightly pressure dependent. Also, the reality is that V in 2) is greater than in 4) because of the higher elevation. Another problem: atmospheric pressure does work on the balloon as you move it down, compressing it.

To make this easier to deal with, the creator of this problem said to make it a Mylar balloon such that when the water is all vapor, it is fully expanded to volume V_o, and then it does not expand any more. We might as well start with temperature at the boiling point that it will have at height h. Now raise it, boil it, then heat it some more until the pressure is P_o . Now pressure, volume and temperature are the same as they will be at the ground level, before condensing the water. Next, pull it down to ground level. Condense it, then lower its temperature to the original temp.

OK,

1) mechanical work E₁ = mgh

2) heat added E₂ = Lm + PV_o + mc_VDT where c_V is the specific heat capacity of the vapor at constant volume, and DT is the difference in boiling points between ground level and at elevation h. I am making all terms positive.
3) mechanical work E₃ = (P_o – P)V_o – mgh to pull it down to ground level.

4) heat removed E₄ = -L_o m – P_oV_o - cmDT where c is specific heat capacity of water, and remember DT is positive, the absolute value of the temperature change. After step 4, the water is identical to its condition at the beginning.

The total is + mc_VDT - cmDT + Lm - L_om = m[(c_V – c)DT +L-L_o]

c is 1.0072 cal/(g ^oC) = 1.0072 BTU/(lb ^oF) for water at the boiling point. The author of the problem tells me that c_v = 0.3352 BTU/lb. He also gives some data, from steam tables, for 1 pound of water raised to 200 ft in this problem:

Heat to boil it at 200 ft: 1150.36 BTU at 14.592 psi and 211.6 ^oF.

Heat to boil it at 0 ft…. 1150.50 BTU at 14.696 psi and 212 ^oF ( so DP = 0.104 psi = 14.976 lb/ft²)

V_o = 26.799 ft³

With the above data, L-L_o = 1150.36 – 1150.50 + PV_o – P_oV_o = -0.14 + 0.52 = 0.38 BTU/lb. (1 BTU = 777.9 ft lb)

(c_v-c)DT = -(0.672)(0.4) = -0.27 BTU/lb

This makes the sum of the E = 0.11 BTU per pound, pending better numbers. Note that a key step is the difference between two almost identical 6 digit numbers, and all the other quantities are good to fewer digits, including a DT to 1 digit precision, making the calculations suspect (GIGO, or more accurately, good data in, garbage out), but I think he will say that this result is evidence against the conservation of energy principle. SE = 0 by energy conservation, SE = 0.11 BTU by the data. But imagine that the 1150.36 BTU/lb is high by a hundredth of a percent, and the correct figure is 1150.25. Then SE = 0. More likely, it is a combination of little errors. We have no exact quantities; there is “noise” in the data, and I think it is safe to say that conservation of energy is in no way falsified by this result.

PV diagram shows that total work by the fluid is negative, thus it is a heat pump, drawing in heat from the atmosphere at low P and expelling heat at the higher P. (Area of the rectangle is about 0.104 lb/in² ∙ 144 in²/ft²∙ 27 ft³ = 404 ft∙lb. This is the work done on the gas.)

14.696 psi

14.592 psi

…………………………………………27….V, ft³

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