Latent Heat: Fusion vs Vaporization (Expert Level)?

Question

Hollow iron sphere full of 1,000 liters of liquid sodium @400C. The iron sphere has an initial average temperature of 50C, and a total heat capacity of 10,000kJ/C. The sphere is submerged in 10,000 liters of water @25C. How much water will flash to steam in a room at a vacuum of 25" Hg?

http://s223.photobucket.com/albums/dd78/floodtl/?action=view¤t=LatentHeat.jpg

http://www.speclab.com/elements/sodium.htm

Assume the air ejector maintains the vacuum as the water flashes to steam.

Doug: The sodium will end up fusing. A percentage of the water will vaporize. That's the question. How much water will vaporize?

m w: Read it again.

The iron sphere has a specific heat capacity of 10,000kJ/kg. That only applies to the iron sphere. I basically just gave you the mass of the sphere x the specific heat capacity of iron. It has nothing to do with the liquid sodium.

Correction: The iron sphere has a TOTAL heat capacity of 10,000kJ/kg.

Answer 1

the way your problem is stated is this....

you have an iron sphere + liquid sodium. That system has a heat capacity of 10,000 kj/C and is at an average temp of 50C.

you have water at 25C under vacuum. which lowers it's boiling point to 56C.

you mix the two and the final temp of the system would be this....

ΔE sphere + ΔE water = 0

10000 kj/C x (1000 cal / 4.184 kj) x (50- Tf) + 10, 000 L x (1000 g / L ) x (1 cal / g C) x (Tf - 25) = 0

(50- Tf) = 4.184 x (Tf - 25)
50 + 104.6 = 5.184Tf

Tf = 29.8 C

no water flashes

you may want to check your problem definition....

******** update ********
jack. i read it again. Still ambiguous. What do you mean by initial? Is the iron sphere at 50C before the addition of the sodium or is it at 50C while being dropped into the water? If I assume it's 50C at the time it's submerged (which is hard to image an iron ball at 50C filled with 400C metal). then.....

ΔE sphere + ΔE sodium + ΔE water = 0

the sphere heats up according to
ΔE sphere = Hc x dT = 10000 KJ/C x (56.11 - 50) = 6.11 x 10^4 KJ

the sodium cools like this.
ΔE liquid sodium + ΔE fusion of sodium + ΔE solid sodium

ΔE liquid sodium = m x cp x dT = 1000 L x (927 g / L ) x Cp liquid sodium x (400C - 97.8C) = 927,000 g x 302.2C x Cp liquid sodium = 927,000 g x 302.2 C x (.331 cal / gC - from ref. below) x (4.184 J / cal) = 3.88 x 10^8 J = 3.88 x 10^5 KJ

ΔE fusion of sodium = m x Hf = 1000 L x (927 g / L) x (1 kg / 1000 g ) x (113 KJ / kg) = 1.05 x 10^5 KJ

ΔE solid sodium = m x Cp x dT = 1000 L x (927 g / L ) x (.292 cal / g C) x (97.8C - 56.11C) x (4.184 J / cal) = 4.72 x 10^7 J = 4.72 x 10^4 KJ

the water heats like this....
ΔE water heating up to it's boiling point + ΔE water as it boils.

ΔE water heating up = m x Cp x dT = 10,000 L x (1000 g / L ) x (1 cal / g C) x (56.11C-25C) x (4.184 J / cal) = 1.30 x 10^9 J = 1.30 x 10^6 KJ

water evaporating = m x Hv = m x 2260 KJ / kg (from second reference attached)

sum it it and you have this.

heat lost = 3.88 x 10^5 KJ + 1.05 x 10^5 KJ + 4.73 x 10^4 KJ = 5.40 x 10^5

heat gained = 6.11 x 10^4 KJ + 1.30 x 10^6 KJ + m x 2260 KJ / kg = 1.36 x 10^6 KJ + m x 2260 KJ/kg

since heat gained > heat lost without any vaporization, the actual final temp is less than the boiling point of the water and no water vaporizes. if I run the calculations with Tf instead of 56.11 C, then....

I get Tf = 40.7C. again no water flashes over



********** another update **********

please check the numbers and units one more time.

you just gave the total heat capacity of the iron sphere as 10,000 KJ / kg. what is the mass of the sphere? you gotta know that... or it has to be in KJ / C and are you sure about that starting temperature? and that 10,000 Liters of water?

if it was 1000 liters of water instead of 10,000 L, it works out to 154 kg water flashes.

Answer 2

Nothing like an easy question, is there? ☺
First, calculate the boiling point of water at 25" Hg. (or look it up)
Then calculate the amount of energy the sphere must lose (at 10,000 kJ/C) to reach that temperature.
Then calculate the amount of energy needed to move 1 liter of water from 25 C to the boiling point and then add the latent heat of vaporization (for 1 liter) to that. (Not sure where latent heat of fusion got in there, you don't make the liquid <-> solid phase transition.)
Now divide that number into the amount of heat the sphere must lose to drop to the boiling point and that's the number of liters that will turn to steam.

Doug

EDIT: Oops..... I did that assuming that the pressure in the room remains at 25" of Hg. If it's a closed room and the BP of the water goes up as it's converted to steam then it -does- become a nasty problem

Answer 3

This equation is not quite correct: ΔE sphere + ΔE sodium + ΔE water = 0 Lets make extra assuptions: Iron sphere is rigid and expands by negligible amount. Pressure (of sodium) inside the sphere is very low. Then, correct conservation of energy is ΔE sphere + ΔE sodium + ΔE water = work ΔE sphere + ΔE sodium + ΔE water = -PΔV water ΔE sphere + ΔE sodium + ΔH water = 0 Latent heats of evaporatation quoted in the books are not ΔE's, they are ΔH's (i.e. enthalpies). ***** The energy is lost when water evaporates an thus expands against ambient pressure.