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#1
10-05-2009, 06:41 PM
How many watts does it take to freeze a gram of water?

With a fairly modern, fairly efficient US ice maker how many watts does it take to cool a gram of water from tap temperature to ice that is average beverage temperature ice?
#2
10-05-2009, 06:54 PM
 _xiao_wenti_ Guest Join Date: May 2002 Location: SE USA Posts: 404
Cp[water, liquid]((T, room in kelvin)-(T, ice in kelvin)) will give you the number of joules needed. Joules are a measure of energy, watts are a measure of power (1 joule per second).

4.186J/gram°C*(21°C-0°C), so ~88J per gram water

You really can't use a time-embedded unit very easily, heat transfer, kinetics, etc will confound it all.
#3
10-05-2009, 06:57 PM
 beowulff Member Join Date: May 2001 Location: Scottsdale, more-or-less Posts: 16,207
Cooling 1g of water from 25 C to 0 C takes 25 calories. Freezing 1g of water at 0 C takes 80 calories, a total of 105 calories.

1 Watt is 859 calories/hour, so at 100% efficiency, 1 Watt could freeze 8.1 grams of water per hour. Figure on a refrigerator being around 30-40% efficient, so say 2-3 grams.

Note that Watts are a unit of power - what you are looking for is watts-hours, which is a unit of energy.
#4
10-05-2009, 06:59 PM
 _xiao_wenti_ Guest Join Date: May 2002 Location: SE USA Posts: 404
Gotta love thermodynamics!
#5
10-05-2009, 08:11 PM
 GameHat Guest Join Date: Dec 2007 Location: NW Chicago Suburbs Posts: 2,363
Quote:
 Originally Posted by _xiao_wenti_ Gotta love thermodynamics!
Spoken like a person who never had to take three semesters of thermodynamics!

Hate, hate, hate it with a passion. Thermo is the engineering equivalent of Flagellantism.

Well, thermo and diff EQs.
#6
10-05-2009, 08:35 PM
 _xiao_wenti_ Guest Join Date: May 2002 Location: SE USA Posts: 404
Two semesters of physical chemistry, two semesters of thermo for metallurgists, one graduate thermo class, one undergraduate class on transport, one graduate class on precipitates, on graduate class on phase diagrams, on graduate class in electrochemistry (corrosion) and lots of other classes touched on it. 4.0-ed almost almost all of it.

Oh, and one of my 4 qualifier sections was thermodynamics.

Oh, yes indeed, I took my dose of thermo. As a materials engineer, thermo is at the root of nearly everything.
#7
10-06-2009, 02:50 AM
 treis Guest Join Date: Jul 2001 Posts: 9,264
Quote:
 Originally Posted by _xiao_wenti_ Two semesters of physical chemistry, two semesters of thermo for metallurgists, one graduate thermo class, one undergraduate class on transport, one graduate class on precipitates, on graduate class on phase diagrams, on graduate class in electrochemistry (corrosion) and lots of other classes touched on it. 4.0-ed almost almost all of it. Oh, and one of my 4 qualifier sections was thermodynamics. Oh, yes indeed, I took my dose of thermo. As a materials engineer, thermo is at the root of nearly everything.
I think its time for a thermo booster shot. You neglected the energy required for the phase change from liquid water to solid ice. Furthermore, what you calculated is the amount of energy removed from the water. This is not the same as the energy required to run the refridgerator.

A refridgerator is different from, say, a stove. A stove provides all of the energy required to heat something up. A refridgerator, on the other hand, is a heat pump. It works by coaxing heat to move from the inside of the refridgerator to the outside, i.e. your kitchen. Your refridgerator is running a vapor compression cycle which, IIRC, has a coefficient of performance in the high 3s. In other words, for every one Joule your refridgerator uses it transfers 3 joules of energy from the inside of the refridgerator to your kitchen.
#8
10-06-2009, 09:40 AM
 robby Charter Member Join Date: Dec 2000 Location: Connecticut, USA Posts: 5,259
Quote:
 Originally Posted by treis I think its time for a thermo booster shot. You neglected the energy required for the phase change from liquid water to solid ice. Furthermore, what you calculated is the amount of energy removed from the water. This is not the same as the energy required to run the refridgerator. A refridgerator is different from, say, a stove. A stove provides all of the energy required to heat something up. A refridgerator, on the other hand, is a heat pump. It works by coaxing heat to move from the inside of the refridgerator to the outside, i.e. your kitchen. Your refridgerator is running a vapor compression cycle which, IIRC, has a coefficient of performance in the high 3s. In other words, for every one Joule your refridgerator uses it transfers 3 joules of energy from the inside of the refridgerator to your kitchen.
Exactly. Also, for all of you posters before treis, you are being sloppy in your terminology.

You have to remove heat to cool and freeze liquid water. Referring to the "number of joules needed" or stating that cooling "takes 25 calories" and that freezing "takes 80 calories" does not make much sense.

To effect the cooling and freezing, the refrigerator's heat pump does indeed require energy, but realize too that no heat pump is 100% efficient.

P.S. I also took two semesters of physical chemistry and two semesters of thermodynamics and learned to hate the whole subject with a passion. The worst part for me were the thermodynamic equations involving partial derivatives.
#9
10-06-2009, 12:23 PM
So let toss the original question and offer a restated new question: how many kilowatt hours does the US use just making ice for beverages?
#10
10-07-2009, 12:10 AM
 treis Guest Join Date: Jul 2001 Posts: 9,264
Quote:
 Originally Posted by janeslogin So let toss the original question and offer a restated new question: how many kilowatt hours does the US use just making ice for beverages?
I actually think it is going to be a negligible amount. I'm going out on a bit of a limb here, but unless an expert comes along I think this is the best you can get. Anyways, at steady state the amount of stuff in your fridge has a negligable effect on the amount of energy. In other words, once everything in the fridge/freezer gets to the set temperature of the fridge it doesn't matter if there is nothing in there or its jam packed. It will use the same amount of energy. Thus, the only difference in energy use is when stuff is being cooled down to the set point of the fridge.

The heat of fusion for water is 334 KJ/KG and the specific heat is 4.186 KJ/KG/C. Thus to freeze 1 KG (1 Liter) of water at 20 c it takes 417.72 KJs. Thats .12 kWh. It looks like the average home uses 936 kWh. So if you freeze 1 liter of water every day, you would use 3.6 kWh per month or .35% of a typical home. In other words the amount of energy used to freeze ice for drinks is basically nothing in the grand scheme of things.
#11
10-07-2009, 12:57 AM
 treis Guest Join Date: Jul 2001 Posts: 9,264
Man, after pointing out the mistake I do the same thing. I forgot to work in the coefficient of performance for the fridge. Its probably somewhere between 3 and 4, which make teh final number 1.2 kWh and .15% of energy usage.

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