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#2624357 - 09/07/08 08:05 PM
Re: Geek question for TFP
[Re: JDavis7873]
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TFF Celebrity
Registered: 02/13/03
Posts: 7772
Loc: Tyler, Texas
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JD,
I'm going to deal with pure H2O here.
Water exists in three different states, solid, liquid, and vapor(gas).
What happens in that when you add heat to ice the temperature of the ice will rise one degree Celsius for each Joule added per kilogram of ice.
So for 10 kg of ice at -40 degrees Celsius adding 10 Joules of thermal energy will raise that 10 kg of ice to -39 degrees Celsius. Adding 100 Joules of thermal energy will raise the 10 kg of ice to -30 degrees Celsius, and so on.
Every thing continues is this fashion till one gets to 0 degrees Celsius. What happens at this point is that the ice starts undergoing a phase transition. This phase transition requires energy to take place. After all the ice has transformed into liquid water then one returns to the original addition of heat to raise the temperature which continues till the liquid water gets to 100 degrees Celsius at which point another phase transition takes place. This is the transformation of liquid water into water vapor. This phase transition also takes thermal energy to take place. Then, once again one goes back to the original addition of thermal energy to raise the temperature of the water vapor.
At the phase transformation from solid to liquid we must deal with what is called the heat of fusion, and at the phase transition from liquid to vapor we must deal with what is called the heat of vaporization. You will often here about something call latent heat. What is being talked about here is the heat that is required to be added to, or taken from water in a particular state to cause that water to undergo a phase transition.
The reason that swamp cooler work so well is because of the heat of vaporization of water is some 2,260 kiloJoules of energy per kilogram. This is a VERY large amount of heat. So when water evaporates it must absorb 2.26 Mega-Joules of heat from its surroundings for each kilogram of water that undergoes evaporation.
Now, contrast this with the heat of fusion of water which is some 335 kiloJoules of energy per kilogram. This is a not such a large amount of heat as compared to the heat of vaporization. In fact is only 0.148 as much thermal energy, or about 15% of the thermal energy needed for vaporization.
Finally, it takes 4.184 kiloJoules of thermal energy to raise one kilogram of water one degree Celsius.
All this means that while you will get additional cooling with colder water that you would with warmer water the amount of additional cooling that you will get will not be nearly as much extra as one might suppose at first blush.
TFP
_________________________
"Two things hold in the greatest of circumspection; government, and self."   
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#2624362 - 09/07/08 08:07 PM
Re: Geek question for TFP
[Re: FishingPhysicist]
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TFF Celebrity
Registered: 05/07/04
Posts: 8154
Loc: Kemp/Waco, Tx
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eeeeeeeeeeerrrrrroooommmmmm
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#2624644 - 09/08/08 04:31 AM
Re: Geek question for TFP
[Re: JDavis7873]
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TFF Celebrity
Registered: 11/09/04
Posts: 7128
Loc: Frisco, Tx
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#2624797 - 09/08/08 05:49 AM
Re: Geek question for TFP
[Re: FishingPhysicist]
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TFF Guru
Registered: 01/31/05
Posts: 21293
Loc: McKinney, TX
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JD,
I'm going to deal with pure H2O here.
Water exists in three different states, solid, liquid, and vapor(gas).
What happens in that when you add heat to ice the temperature of the ice will rise one degree Celsius for each Joule added per kilogram of ice.
So for 10 kg of ice at -40 degrees Celsius adding 10 Joules of thermal energy will raise that 10 kg of ice to -39 degrees Celsius. Adding 100 Joules of thermal energy will raise the 10 kg of ice to -30 degrees Celsius, and so on.
Every thing continues is this fashion till one gets to 0 degrees Celsius. What happens at this point is that the ice starts undergoing a phase transition. This phase transition requires energy to take place. After all the ice has transformed into liquid water then one returns to the original addition of heat to raise the temperature which continues till the liquid water gets to 100 degrees Celsius at which point another phase transition takes place. This is the transformation of liquid water into water vapor. This phase transition also takes thermal energy to take place. Then, once again one goes back to the original addition of thermal energy to raise the temperature of the water vapor.
At the phase transformation from solid to liquid we must deal with what is called the heat of fusion, and at the phase transition from liquid to vapor we must deal with what is called the heat of vaporization. You will often here about something call latent heat. What is being talked about here is the heat that is required to be added to, or taken from water in a particular state to cause that water to undergo a phase transition.
The reason that swamp cooler work so well is because of the heat of vaporization of water is some 2,260 kiloJoules of energy per kilogram. This is a VERY large amount of heat. So when water evaporates it must absorb 2.26 Mega-Joules of heat from its surroundings for each kilogram of water that undergoes evaporation.
Now, contrast this with the heat of fusion of water which is some 335 kiloJoules of energy per kilogram. This is a not such a large amount of heat as compared to the heat of vaporization. In fact is only 0.148 as much thermal energy, or about 15% of the thermal energy needed for vaporization.
Finally, it takes 4.184 kiloJoules of thermal energy to raise one kilogram of water one degree Celsius.
All this means that while you will get additional cooling with colder water that you would with warmer water the amount of additional cooling that you will get will not be nearly as much extra as one might suppose at first blush.
TFP
beat me to it 
_________________________
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#2624804 - 09/08/08 05:51 AM
Re: Geek question for TFP
[Re: TexDawg]
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TFF Celebrity
Registered: 11/09/04
Posts: 7128
Loc: Frisco, Tx
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JD,
I'm going to deal with pure H2O here.
Water exists in three different states, solid, liquid, and vapor(gas).
What happens in that when you add heat to ice the temperature of the ice will rise one degree Celsius for each Joule added per kilogram of ice.
So for 10 kg of ice at -40 degrees Celsius adding 10 Joules of thermal energy will raise that 10 kg of ice to -39 degrees Celsius. Adding 100 Joules of thermal energy will raise the 10 kg of ice to -30 degrees Celsius, and so on.
Every thing continues is this fashion till one gets to 0 degrees Celsius. What happens at this point is that the ice starts undergoing a phase transition. This phase transition requires energy to take place. After all the ice has transformed into liquid water then one returns to the original addition of heat to raise the temperature which continues till the liquid water gets to 100 degrees Celsius at which point another phase transition takes place. This is the transformation of liquid water into water vapor. This phase transition also takes thermal energy to take place. Then, once again one goes back to the original addition of thermal energy to raise the temperature of the water vapor.
At the phase transformation from solid to liquid we must deal with what is called the heat of fusion, and at the phase transition from liquid to vapor we must deal with what is called the heat of vaporization. You will often here about something call latent heat. What is being talked about here is the heat that is required to be added to, or taken from water in a particular state to cause that water to undergo a phase transition.
The reason that swamp cooler work so well is because of the heat of vaporization of water is some 2,260 kiloJoules of energy per kilogram. This is a VERY large amount of heat. So when water evaporates it must absorb 2.26 Mega-Joules of heat from its surroundings for each kilogram of water that undergoes evaporation.
Now, contrast this with the heat of fusion of water which is some 335 kiloJoules of energy per kilogram. This is a not such a large amount of heat as compared to the heat of vaporization. In fact is only 0.148 as much thermal energy, or about 15% of the thermal energy needed for vaporization.
Finally, it takes 4.184 kiloJoules of thermal energy to raise one kilogram of water one degree Celsius.
All this means that while you will get additional cooling with colder water that you would with warmer water the amount of additional cooling that you will get will not be nearly as much extra as one might suppose at first blush.
TFP
beat me to it Ya sure 
_________________________
 
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#2624808 - 09/08/08 05:52 AM
Re: Geek question for TFP
[Re: TexDawg]
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TFF Celebrity
Registered: 03/28/02
Posts: 5567
Loc: BIG "D"...Denton.
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JD,
I'm going to deal with pure H2O here.
Water exists in three different states, solid, liquid, and vapor(gas).
What happens in that when you add heat to ice the temperature of the ice will rise one degree Celsius for each Joule added per kilogram of ice.
So for 10 kg of ice at -40 degrees Celsius adding 10 Joules of thermal energy will raise that 10 kg of ice to -39 degrees Celsius. Adding 100 Joules of thermal energy will raise the 10 kg of ice to -30 degrees Celsius, and so on.
Every thing continues is this fashion till one gets to 0 degrees Celsius. What happens at this point is that the ice starts undergoing a phase transition. This phase transition requires energy to take place. After all the ice has transformed into liquid water then one returns to the original addition of heat to raise the temperature which continues till the liquid water gets to 100 degrees Celsius at which point another phase transition takes place. This is the transformation of liquid water into water vapor. This phase transition also takes thermal energy to take place. Then, once again one goes back to the original addition of thermal energy to raise the temperature of the water vapor.
At the phase transformation from solid to liquid we must deal with what is called the heat of fusion, and at the phase transition from liquid to vapor we must deal with what is called the heat of vaporization. You will often here about something call latent heat. What is being talked about here is the heat that is required to be added to, or taken from water in a particular state to cause that water to undergo a phase transition.
The reason that swamp cooler work so well is because of the heat of vaporization of water is some 2,260 kiloJoules of energy per kilogram. This is a VERY large amount of heat. So when water evaporates it must absorb 2.26 Mega-Joules of heat from its surroundings for each kilogram of water that undergoes evaporation.
Now, contrast this with the heat of fusion of water which is some 335 kiloJoules of energy per kilogram. This is a not such a large amount of heat as compared to the heat of vaporization. In fact is only 0.148 as much thermal energy, or about 15% of the thermal energy needed for vaporization.
Finally, it takes 4.184 kiloJoules of thermal energy to raise one kilogram of water one degree Celsius.
All this means that while you will get additional cooling with colder water that you would with warmer water the amount of additional cooling that you will get will not be nearly as much extra as one might suppose at first blush.
TFP
beat me to it Thanks. I'll keep you in mind when I start working on my redneck nuclear reactor.
_________________________
  The threat of Clowns is real.
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#2624820 - 09/08/08 05:57 AM
Re: Geek question for TFP
[Re: JDavis7873]
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TFF Guru
Registered: 01/31/05
Posts: 21293
Loc: McKinney, TX
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JD,
I'm going to deal with pure H2O here.
Water exists in three different states, solid, liquid, and vapor(gas).
What happens in that when you add heat to ice the temperature of the ice will rise one degree Celsius for each Joule added per kilogram of ice.
So for 10 kg of ice at -40 degrees Celsius adding 10 Joules of thermal energy will raise that 10 kg of ice to -39 degrees Celsius. Adding 100 Joules of thermal energy will raise the 10 kg of ice to -30 degrees Celsius, and so on.
Every thing continues is this fashion till one gets to 0 degrees Celsius. What happens at this point is that the ice starts undergoing a phase transition. This phase transition requires energy to take place. After all the ice has transformed into liquid water then one returns to the original addition of heat to raise the temperature which continues till the liquid water gets to 100 degrees Celsius at which point another phase transition takes place. This is the transformation of liquid water into water vapor. This phase transition also takes thermal energy to take place. Then, once again one goes back to the original addition of thermal energy to raise the temperature of the water vapor.
At the phase transformation from solid to liquid we must deal with what is called the heat of fusion, and at the phase transition from liquid to vapor we must deal with what is called the heat of vaporization. You will often here about something call latent heat. What is being talked about here is the heat that is required to be added to, or taken from water in a particular state to cause that water to undergo a phase transition.
The reason that swamp cooler work so well is because of the heat of vaporization of water is some 2,260 kiloJoules of energy per kilogram. This is a VERY large amount of heat. So when water evaporates it must absorb 2.26 Mega-Joules of heat from its surroundings for each kilogram of water that undergoes evaporation.
Now, contrast this with the heat of fusion of water which is some 335 kiloJoules of energy per kilogram. This is a not such a large amount of heat as compared to the heat of vaporization. In fact is only 0.148 as much thermal energy, or about 15% of the thermal energy needed for vaporization.
Finally, it takes 4.184 kiloJoules of thermal energy to raise one kilogram of water one degree Celsius.
All this means that while you will get additional cooling with colder water that you would with warmer water the amount of additional cooling that you will get will not be nearly as much extra as one might suppose at first blush.
TFP
beat me to it Thanks. I'll keep you in mind when I start working on my redneck nuclear reactor. i'll sharpen my crayons
_________________________
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#2624935 - 09/08/08 06:46 AM
Re: Geek question for TFP
[Re: FishingPhysicist]
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TFF Celebrity
Registered: 07/21/05
Posts: 5888
Loc: Ft. Worth
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TFP if I may without too much insult, I'm going to translate this for the less scientificly educated. JD,
I'm going to deal with pure H2O here.
Water exists in three different states, solid, liquid, and vapor(gas).
What happens in that when you add heat to ice the temperature of the ice will rise one degree Celsius for each Joule added per kilogram of ice.
BLAH BLAH BLAH
BLAH BLAH BLAH
BLAH BLAH BLAH
BLAH BLAH BLAH
BLAH BLAH BLAH
BLAH BLAH BLAH
All this means that while you will get additional cooling with colder water that you would with warmer water the amount of additional cooling that you will get will not be nearly as much extra as one might suppose at first blush.
TFP
_________________________
FATTY YOU SADISTIC SOB! BUY THAT DOG 64  Did you eat an extra bowl of stupid this morning?
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#2624955 - 09/08/08 06:53 AM
Re: Geek question for TFP
[Re: Kingfisher196]
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TFF Celebrity
Registered: 04/18/02
Posts: 5516
Loc: Irving, Tx
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TFP if I may without too much insult, I'm going to translate this for the less scientific educated. JD,
I'm going to deal with pure H2O here.
Water exists in three different states, solid, liquid, and vapor(gas).
What happens in that when you add heat to ice the temperature of the ice will rise one degree Celsius for each Joule added per kilogram of ice.
BLAH BLAH BLAH
BLAH BLAH BLAH
BLAH BLAH BLAH
BLAH BLAH BLAH
BLAH BLAH BLAH
BLAH BLAH BLAH
All this means that while you will get additional cooling with colder water that you would with warmer water the amount of additional cooling that you will get will not be nearly as much extra as one might suppose at first blush.
TFP
Cool! I understand now!
_________________________
"If a man is truly blessed, he returns home from fishing to be greeted by the best catch of his life" 
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