Physics Thermal Energy Question?

Question

I own a large swimming pool. It is 20FT wide, 40FT long, and has an average depth of 5FT. The top 2FT of pool is frozen solid (avg ice temp 24F, water temp avg below ice is 34F). Pool is enclosed so that outside conditions do not affect the pool environment. I rented a magic heater to run for 35 hours, dumping heat into the pool at a rate of 292,000 cal/sec. After 36 hours, is it safe to go swimming? what is the condition of the water at this time - e.g. half ice, half water, all water, some steam, what teamptures? Please show all your work.

Assumptions: 30.48 cm per foot
density of ice = 0.88 g/cc
density of water = 1 g/cc
specific heat of ice == 0.5 cal,g-c
specific heat of water = 1.0 cal/g-c
heat of fusion for water/ice = 80 cal/g
heat of vaporization water/steam = 540 cal/g

heater is run for 36 hours, not 35.. my bad.

Answer 1

OK, first of all
20 ft * 30.48 cm/ft = 609.6 cm
40 ft * 30.48 cm/ft = 1219.2 cm

2 ft * 30.48 cm/ft = 60.96 cm
3 ft * 30.48 cm/ft = 91.44 cm

So, we have 609.6 cm * 1219.2 cm * 60.96 cm = 45306954.5472 cc of ice at 24F and
609.6 cm * 1219.2 cm * 91.44 cm = 67960431.8208 cc at 34F

24F = -4.4C
34F = 1.1C

36 hours * 3600 sec/hr * 292,000 cal/sec = 37843200000 cal generated by the heater.

First we are going to bring the ice to the melting point.

45306954.5472 cc * .88 g/cc * 0.5 cal/gC * 4.4C = 87714264.0033792 cal

37843200000 cal - 87714264.0033792 cal = 37755485735.9966208 cal remaining, so we can do more work.

Let's convert the ice to water.

45306954.5472 cc * .88 g/cc * 80 cal/g = 37755485735.9966208 cal - 3189609600.12288 cal = 34565876135.8737408 cal, so we can do more work.

Let's heat this water to 1.1 C to bring it to the same temperature as the rest of the water.

45306954.5472 cc * .88 g/cc * 1 cal/gC * 1.1 = 43857132.0016896 cal

34565876135.8737408 cal - 43857132.0016896 cal =
34522019003.8721 cal, so we can do more work.

Now, let's consider all of the water and start heating it.

67960431.8208 cc of water that was always at 1.1 C
67960431.8208 cc * 1 g/cc = 67960431.8208 g
45306954.5472 cc * .88 g/cc = 39870120.001536 g

Add them together and get 107830551.822336 g

107830551.822336 g * 1 cal/gC = 107830551.822336 cal/C

Let's divide 34522019003.8721 cal / 107830551.822336 = 320 C. Well, long before that, it's going to start boiling, so that won't work.

The water is at 1.1 C and water boils at 100 C so 100 - 1.1 = 98.9 C.

107830551.822336 g * 1 cal/gC * 98.9 C = 10664441575.2290304 cal

34522019003.8721 cal - 10664441575.2290304 cal = 23857577428.643 cal remaining, so we can do more work (although we already knew that).

So let's start boiling the water.

107830551.822336 g * 540 cal/g = 58228497984.06144 cal

23857577428.643 cal - 58228497984.06144 cal = --34370920555.4184 cal, so we don't have enough calories to convert all of the water to steam.

23857577428.643 cal / 58228497984.06144 cal = 0.4097 = 40.97% of the water will have converted to steam. 59.03% of the water will remain.

107830551.822336 g * 59.03% = 5490731698.79334912 g of water will remain. That is 63652374.740725 cc

Going back to our original problem, 63652374.740725 cc / (609.6 cm * 1219.2 cm) = 85.64 cm = 2.8 feet.

So, this amount of heat is sufficient to melt all of the ice and turn it and 20% of the liquid water to steam. What is left is a pool of water 2.8 feet deep that is at boiling temperature.

DON'T GO SWIMMING.

Answer 2

Mass of water = volume of water * densitywater

Mass of ice = volume of ice * densityice

Of course, convert those feet to cm so your density calculation works.

The energy you are dumping (E) is power (P) * time (t)

First get everything to 0C by raising the ice by 8F (convert that!) and lowering the water be 2F. Lowering the water will give you heat. Raising the ice will take it. Heat given/taken = temp change * mass * specific heat. Subtract the heat taken from your total.

energy to melt ice is massice*heat of fusion

subtract that energy from the running total

The energy to raise that to 100 degrees C is (massice+masswater)*100 degrees*specheatwater

The energy to boil that is (massice+masswater)*heat of vaporization

So given how much energy you have, figure out how much that does (melts the ice?, heats the water to 100 also? boils the water also? heats the steam some more?)

Then you can figure out the temperature (if you run out of energy during a heating phase) or the percent boiled/melter (if you run out of energy during a phase change)

delta temp = remaining energy / (mass * specific heat)
OR
percent phase-changed = energy remaining / energy required to phase-change everything

Have fun with that!

Answer 3

Efficiency will increase, of course. But remember, you cannot decrease the temperature indefinitely.