# Creating water

1. Apr 13, 2007

### Papak

Hi again,

I was doing some experiments on heat of combustion of various fuels and determine which fuel is better by creating a calorimeter. And its simple and straight forward but i started thinking what happens to the water created by burning the fuel, as an example i was doing methanol and from the formula i get 1.05grams of water for the period of time i ran the experiment and from the amount of fuel burnt. I was thinking where is it created, where does it go what temperature will it be and how will this affect my results. I've been thinking about this for some while and am kinda confused and interested about this, it is irrelevant to my experiment but still i want to know more about it. I checked hundreds of websites and they all dont say much about what i want and my text books are like prehistoric and really really vague, Hoping someone can shed some light on my question. Thanks in advance

Regards Papak

2. Apr 14, 2007

### ShawnD

My gut tells me it's probably not a big deal since gaseous water is nothing special. Gaseous water does not have high surface tension, it does not have high heat capacity, and it's not extraordinarily light or heavy. Basically it's no different from CO2, O2, or N2 when it's a gas, so it shouldn't have any affect on your calorimeter results.

I also think most heat of combustion numbers factor in any effects of water, so seeing something like X amount of energy for gasoline and 1.3X amount for diesel is a representation of usable energy you get from burning them.

3. Apr 14, 2007

### Papak

thank you very much, i also did a few extra calculation including this extra water and it really does nothing to my final results (i used a very wide range of varibles for temperature) so thanks again for the help.

Regards Papak

4. Apr 15, 2007

### JPRitchie

See the wiki entry for heat of combustion and calorimeter for basic information, as well as other common thermodynamic quantities. Water does frequently have anamolous properties, because of hydrogen bonding. For example, the wiki has some values for surface tension (gases don't have surface tension). Water has a much higher value compared with nonane, methanol, heptane, and ether, for example. Water has a heat of vaporization (near 1 atm.) similar to that of methanol ( ca. 41 vs. 37 kJ/mol, respectively), but higher than non-hydrogen bonded liquids such as propane ( 15.7) and butane (21.0). Whether this is significant or not can depend upon the molecular formula of the fuel and the final p,T of the calorimeter. Finally, remember that heat liberated isn't useful work to, for example, drive an internal combustion engine; expansion of gases provides that.
-Jim

5. Apr 15, 2007

### Staff: Mentor

The heat of combustion data for fuels is tabulated and the effect of water in the combustion products can be on the order of about 15% depending on whether the water is condensed out or left as a vapor. Any half decent thermodynamics textbook will have a couple of chapters on this and tables of the chemical energies involved.

6. Apr 18, 2007

### Papak

Hehehe well neither did it make much sense in my mind but neway. I checked wikipedia and all and it really didnt help much with what i was looking at becuase im only doing High shool Chemisty, i know how an internal combustion engine runs but i wast looking at fuels only for running cars or engines but looked at the best fuel in relation to the energy released during combustion in joules. What i was wondering and what i asked was whether the water created in combustion would alter the actual value of these fuels by taking in some amount of energy or somehow other alter my final results. But as ShawnD said it was more likely pointless as it was very slight. I only got a full 1.05g of water produced and changed a lot of varibles such as the Specific heat capacity, the inital temp and final and the amount of heat it could absorb was minimum. And thank you russ_watters ill borrow a thermodynamic book next time im at the library and read more into it, its seems quite interesting. Regards papak

7. Apr 18, 2007

### pixel01

The product water state is very important in determining the heating value of a fuel. If the water created is liquid, you have the high heating value or HHV. If it is vapor, you have low heating value - LHV. The difference between HHV and LHV is the vaporazation of water at 100 degrees C which is 2.3kJ/gram

8. Apr 18, 2007

### Papak

Im unsure what you are saying pixel. does that mean that for every gram of liquid water created, 2.3kj's of energy given off from the combustion reaction will be taken in by the water, making my actual answer 2.3kj*amount of water in g, less than what it should be or is it the opposite. Regards Papak

9. Apr 19, 2007

### pixel01

No, contrarily, for every gram of liquid water created, you have 2.3kj more compared to the case that amount of water is still in the gas phase. Remember, the vapor water gives off heat when it condenses.

10. Apr 19, 2007

### Papak

Ah okay well thats much better becuase ive only had condensed water on one of the experiments thank you for clarifying this and now that you put it that way its very logic and i feel foolish for not realizing it myself straight away hehe. Thanks to every1 for helping

11. Apr 19, 2007

### pixel01

It is nice to hear that. I think that is the way we help each other in the forum.

12. Apr 22, 2007

### Papak

Hehe i guess and thank you to all again, its lead to some interesting reading.