Molten Gold Droplets in Water

Do an energy balance on the gold-water system, assuming:

1. No chemical reaction
2. The gold and the water achieve the same temperature
3. No thermal loss to surroundings

Your best bet is to send the whole lot to me here in

and I'll work it out for you. <Splutter>

Good answer GB.

STEAM!

milo

Definitely, you've a good answer! there are two stages took places must be consider. On immediates heat lost on gold,as heat radiating (Crazzy sizzling! its delta temp) through the water- to deter on how long on the bulk of water is becoming equilibrium and water volume after vaporated in total. My randomly guess IS NOT 5° rise in temp.

Phase 1 gold temperature >> water boiling temperature. heat is used to vaporize water so that one has to consider the latent heat to transform water in vapor+ heating of water to its vapor temperature.

Phase 2 gold temperature < water boiling temperature normal heat transfer without phase change.

In fact your thought was correct but only for one component since you considered the latent heat for gold but forgot that the opposite will happen to water.

The remaining water in the tank will only get the cooling of gold from boiling temperature to the resulting of the energy balance since the rest will go as vapour in the air.

a lot depends on the nature of the stream. They make iron shot by dropping melted iron into water as a rain. Will you rain the gold into the water? or one fast slug-pour?

The fast slug of molten gold might startle the kitties indeed. But if the sped is quick, it might get submerged so quickly that the spatters are trapped and you just get some decrepitation steam evolving and collapsing, much like cavitation.

try googling zinc/copper/gold/iron/steel shot or pellets. You might find some hints.

In any event, you can start with one pound of molten gold to start and work up.

thanks to all for prompt reply. but matter is unresolved. the gold will be poured in the form of rain. 6 balls of 1.5 mm dia at a time.this 6 ball doesnt weigh 50 kg, it is the total amount of gold to be poured. so there will not be anything like splashing. i have not considered latent heat of water because i assume the temperature of water may not go beyond 100 degree c. if i apply latent heat of water(vapourization) then the temperature rise of water is coming negative. and as per my calculation(neglecting latent heat of water) the total rise in temp will be only 5 degrees. pls support

Hi,

your calculation looks good but what are the numbers you did put into the calculation?

5° temp rise seem to be very unlikely!

A similar method is used by traditional Japanese silversmiths.

Pouring molten silver into a pot of water where a sheet of cotton is stretched so that the molten silver at solidifying does not hit the bottom and is some centimeter below the surface. The silver will cool slowly at boiling water around and the material is said to be much better than if solidified in air.

I would suggest a test maybe with silver or copper, start with a small amount and measure: much easier!

RHABE

Just looking at the process that you are proposing and not at the calc side of things, can we assume that the water be continuously changed by a feeder and an outlet. If this were to be the case, then you would probably not exceed 40 or 60 degrees depending on the flow rates. The difference in temp between 1100 degrees and the 20 to 60 degrees variation is probably too small to make any difference.

The feeder would be below water level and the outlet also. This would avoid rippling, if that were a condition.

I like the idea of catching the balls at low level in a net.

Using specific heat capacity of gold to be 0.1291 KJ/Kg/°K (watch out Wikipedia has 0.2292)

specific heat capacity of water to be 4.1813 KJ/Kg/°K

specific latent heat of fusion of gold to be 63 KJ/Kg/°K

I get 37.042°C

But watch out some water will probably turn to steam when you first pour in the gold, this will lower the final temperature.

Whatever happened to "We don't do homework for you"? None of the gold manufactureres do this. Why would you melt the gold bars just to pour it into water to cool it back down? This is only done with an alloy in order to give the alloy a special color or ductile property that can only be achieved by rapid quenching and then it goes in as a hot solid - not a molten liquid. This is a classical homework problem where vaporization of water, which obviously will happen, is ignored or assumed to not happen in order to simplify the problem to a first year algebra-based physics course level, and the student is simply asked to use the equations and concepts they have learned to find a temperature change for one of the materials involved in the heat transfer process. Determining how much steam is produced etc is left for higher level classes unless the heat transfer is enough to raise the temperature of the water to boiling temperature and additional heat is still available.

The mistake he is making is that he is using Tg for temperature of the gold beforehand and Tw for temperature of the water after. These should both be delta Ts or the change (delta) in the temperature of the gold and the change in the temperature of the water. A simple referral to the text book or a look at his class notes - even a look it up on google would have provided this information and he could have had the satisfaction of finding the correct answer himself.

well, if you sell gold by the gram, droplet shaped bits might make them easy to pour and weigh.

Will it form smooth drop shapes or some tangled mass.

Easy to experiment with melted tin or solder.

"Why would you melt the gold bars just to pour it into water to cool it back down? "

Hi,

as the spheres have a diameter of only 1.5mm there may be more possibilities for need and use:

Etruscan style jewelry: soldering or welding of small spheres in ornaments to base structure, both 999-pure gold?

Other more modern style jewelry?

Contacts for high reliability high current switching?

Ingots for precision alloying - dental use?

I am sure some more existing.

RHABE

Profs have many times a view concentrated on school and forget the "artistic" realm.

The tendency to make jewels with sintered gold spheres is very actual today. I think it is the goal of the person who asked the question.

Agreed.

however, 50kg is a lot of mille-spheres!

milo

You cannot imagine the quantities in series production.

thanks to all for providing me the solution. RGO i followed your formula and found out the temperature rise as 25 degrees, and which is correct. can you please tell me what values you have taken for latent heat, specific heat etc. bye the way i have taken mass of water= 340 kg mass of gold= 50 kg cp of water=4.18 j/g/k laten heat of gold= 67.5 j/kg cp of gold=0.128 j/g/k. bye the way i am an engineer.

The specific and latent heats he has used are in the links he has provided, and, if you plug those numbers straight into his formula you get

T_f = 37.066 °C

Not that it will make a lot of difference to the result, but, where did you get the latent heat of fusion of gold to be 67.5. KJ/Kg. Both our sources had 63KJ/Kg.

C_gold=0.03kcal/kg.K

L_gold=15kcal/kg

c_water=1kcal/kg.K

Final temperature of both water and gold ≈ 37 ºC

By the way, is this a real problem?

Glad to be of help.

Cheers

yeah RGO it is the real problem, we are the designer and suplier of induction furnaces and one of our customer in saudi wants to make golden ball, for which he asked us to provide the rise in temperature. he has asked that the temperature should not be more than 70 degrees celcius. we have to provide them a water tank in which they will pour golden drops. my questions looks very simple and i accept it is very simple question but why i used this forum is just because i had some doubt in using the latent heat of gold and water. just because of that i asked this stupid question to you and all. thanks to physicsprof, randall RGO and others for timely help. bye

If you want perfect gold spheres: perhaps you should consider a shot tower, or, a modification of this method:-

http://cr4.globalspec.com/comment/360377

Water boils at 1100°C at a pressure of about 34 atmospheres so you'll need a stout container.

Or you could try the Bliemeister method mentioned at post #43 in the same thread.

Hello manas_dr

I am wondering why you would want or need to pour so much gold at one time?

If I understand you correctly, you have seven times as much water as gold. The gold is going nowhere, but, the water will almost explode. You really need to do this with some sort of body protection or remotely.

Can you say what you are making? As if you expect the gold to turn to droplets, the droplets will differ in size depending on the heat of the water.

Take care and please tell us what happens.

hello babybear as i told earlier that we are the furnace provider and not the sperical ball maker. actually 50 kg is not poured at a time. it is a process of hours. i am just checking whether the temperature rose above 70 degree c if i poured 50 kg at once. actually gold is poures in form of 7 sperical ball at a time and the diameter is 1.6 mm only, so at a single operation hardly a kg of gold is poured.

Hello manas_dr,

Now that explanation sounds believable ! A friend used to do the same thing linked to jewelry, and never poured more than 100 grams. I was wondering who was rich enough to be able to pour 50 kg in one go?

Take care.....

Another way to deal with this is to drop the gold into another metal that is molten above the melting point of gold but is held below the melting point of the gold, so the falling gold will form spheres as it falls = no boiling, spattering etc. It should be a metal with which gold does not form an alloy or other mixture or compound.

You could also drop them in vacuum in an evacuated drop tower with oil at the base. Choose the height and starting temperature of the gold so it will be strong enough by the time it hits the oil that it will not deform, start a little above the melting point.

You can probably use a heated pot with a tapered hollow cone and valve of the right diameter to give you the drop size you need and open the valve a little and the stream will form drops and they will fall off and shape/cool

Hi manas_dr,

I know this was a 'hyp', but that much Gold would be worth at least £800,000, almost $122,000,000!............Rough estimate!!!!!

Take care

I need £400,000 changed, send me $61,000,000 and keep $1,000,000 for ur trubble

Got the figures wrong there.!

Should be: 1.22 million

Sorry