Friday, June 7, 2013
Introduction to Paper Mill Sludge Problem and Proposed Idea
The paper mill in Newberg, Oregon needs an environmentally friendly way to dispose of the waste (sludge) that results from their paper making process. Their current method for dealing with this sludge is incineration. The problem with this is that it is expensive and causes pollution. I need to find a way to either:
- Find an environmentally friendly way to dispose of the waste
- Make something of the waste
- Neutralize the odor
- Make the waste less of a fire hazard
I decided that making something of the waste would be the best way to go, because the paper mill might be able to make some money off of what would usually be thrown away. To find out what can be made, I needed to look at the chemical composition of the sludge.
Cellulose (C6H10O5)n 50-80%
Calcium Carbonate CaCO3 8-12%
Silicon Si 2-10%
Aluminum Al <1-10%
The paper mill sludge is mostly composed of cellulose fibers. Cellulose is a natural polymer, with six carbon atoms, ten atoms, and five atoms in each monomer.
Cellulose can be used to reinforce concrete, making it be able to withstand the same tensile stress as carbon and steel microfibers1 . It would be a better alternative to the concrete companies, as it would be cheaper than rebar.
I need to figure out how the paper mill can prepare the sludge to be ready for the concrete making company they would sell it to. Also, I need to find out how the company would use the product in their concrete. Most importantly, I need to see if this would be beneficial to both the paper mill and the concrete company.
1 http://www4.uwm.edu/cbu/Papers/2003%20CBU%20Reports/REP-501.pdf
1 http://www4.uwm.edu/cbu/Papers/2003%20CBU%20Reports/REP-501.pdf
Tuesday, May 28, 2013
Determining Specific Heat of an Unknown Metal
1. What was the purpose of the lab?
The purpose of this lab was to find the specific heat of a metal, for the purpose of finding out which metal it is. We were to find the specific heat by using a hot plate, a calorimeter, water, and an electric thermometer.
2. What did I need to know before undertaking this lab?
The specific heat of a metal is the amount of heat per gram required to raise the substance by one degree Celsius. In this lab, it was crucial to know how to determine the specific heat. We had to start with this equation:
q = mcΔT
In this equation, "q" stands for heat energy gained, "m" means the mass of the substance, "c" means the specific heat, and "ΔT" stands for the change in temperature. Different metals have different specific heats, which we also needed to know.
Substance Specific Heat
Water 4.184
Aluminum 0.897
Brass 0.385
Copper 0.385
Lead 0.129
Stainless Steel 0.490
Zinc 0.390
We couldn't measure the temperature of the metal directly, or know the heat lost from the metal, without using the heat equation for water. All we need to do is solve the equation to find the heat energy gained for the water inside the calorimeter, then plug that in for the equation for the metal to find the specific heat. The heat gained by the water is the same as the heat lost by the metal. This will make more sense when the procedure is explained.
3. What did I do in the lab and what data was collected?
Procedure:
- Take the mass of the coffee cup(calorimeter). Record.
- Pour 100 mL of water into the coffee cup(calorimeter).
- Determine the mass of the water and coffee cup together and record.
- Find initial temperature of the water.
- Fill a beaker with 300 mL of water.
- Place on hot plate and begin getting water to boil.
- Obtain the sample of metal you will be using and measure its mass. Record.
- Place the metal into the beaker and continue heating.
- After 10 minutes, record temperature of the boiling water without the probe touching the inside of the beaker.
- Using tongs, remove metal from beaker and quickly place into coffee cup(calorimeter). Put the lid on quickly, and put the temperature probe inside the calorimeter.
- Wait until temperature is stable and record it.
- Do this procedure at least twice.
Data:
Trial #1:
Mass of Coffee Cup 2.985g
Mass of Coffee Cup with Water 97.7775g
Initial Temperature of Water in Cup 21.1°C
Mass of metal sample 68.205g
Temperature of boiling water 99.2°C
Final Temperature of Water In Cup 27.3°C
Trial #2:
Mass of Coffee Cup 3.397g
Mass of Coffee Cup with Water 92.633g
Initial Temperature of Water in Cup 21.5°C
Mass of metal sample 68.07g
Temperature of boiling water 99.4°C
Final Temperature of Water In Cup 28.1°C
4. What does the data mean? (analysis, calculations)
Trial #1:
- First, we need to plug in the specific heat of water, the temperature change of the water, and the mass of the water, to find the heat gained by the water.
- q = mcΔT
- q= (97.775g-2.985g)(4.184)(27.3°C-21.1°C)
q= (94.79g)(4.184J/g-°C)(6.2)
q= 2458.9J
- Second, we need to find the specific heat of the metal.
- q = mcΔT
- 2458.9J= (68.205g)c(99.2°C-27.3°C)
2458.9J= (94.79g)c(71.9°C)
c= 0.501J/g-°C
Trial #2:
- First, we need to plug in the specific heat of water, the temperature change of the water, and the mass of the water, to find the heat gained by the water.
- q = mcΔT
- q= (89.236g)(4.184J/g-°C)(6.6)
q= 2464.189J
- Second, we need to find the specific heat of the metal.
- q = mcΔT
- 2464.189J= (68.07g)c(71.3°C)
c= 0.508J/g-°C
The average of the two specific heats is 0.5045J/g-°C. This obviously won’t be perfect, because heat could have escaped when we were transporting the metal to the calorimeter from the beaker of boiling water.
5. Conclusion.
After going through the steps to find the specific heat of the unknown metal, we reached the conclusion that the specific heat is approximately 0.5045J/g-°C. This doesn’t fit any of the metals on the chart, but it most closely fits stainless steel.
Monday, April 29, 2013
Decomposition Reaction
Sodium Hydroxide => Sodium Oxide+Water
2NaOH => Na 2 O + H 2 O
To test this we reaction , we had to heat sodium hydroxide by putting it over a candle. We had to see if water formed and if mass was lost.
First we took the starting mass, and it was 63.845g. After a bit, we did clearly see water form.
The resulting mass didn't change as much as we hoped, it only went down to 63.603g.
The Sodium Oxide looked like this, in the end:
2NaOH => Na 2 O + H 2 O
To test this we reaction , we had to heat sodium hydroxide by putting it over a candle. We had to see if water formed and if mass was lost.
First we took the starting mass, and it was 63.845g. After a bit, we did clearly see water form.
The resulting mass didn't change as much as we hoped, it only went down to 63.603g.
The Sodium Oxide looked like this, in the end:
Thursday, February 28, 2013
Writing Formulas for Ionic Compounds
People got confused when they were trying to make ionic compounds from the compound names. It's simple, really, it's just basic multiplication and addition.
_____________________________________________________
For the first example, I'll write the formula for Beryllium Phosphate:
The first thing you should do is write down the ions in the compound name.
Beryllium: Be+2
Phosphate: PO4-3
The second thing you need to do is balance the charges, so they equal zero. The two charges we have at the moment are +2 and -3, which add up to -1, not 0. To make them equal zero, you need different amounts of the chemicals, which would multiply the charges.
Be+2 PO4-3
With this, we would have three beryllium ions and two phosphate ions.
2*3 = 6 -3*2= -6
Making this the formula name for beryllium phosphate:
Be3(PO4)2
See, the charges are nullified. If the parentheses are confusing, remember that they just keep the ion together, and the subscript two after the parentheses means that there are two phosphate ions.
_______________________________________________________
For the second, easier, example, I'll do potassium chlorate:
First, write down the ions in the compound name.
Potassium: K +1
Chlorate: ClO3-1
Again, you have to balance the charges so that they add up to 0.
K +1 ClO3-1
1+-1 = 0
These ions already balance each other out, so you don't have to do anything.
So, potassium chlorate is KClO3 .
________________________________________________________
\if that makes sense
_____________________________________________________
For the first example, I'll write the formula for Beryllium Phosphate:
The first thing you should do is write down the ions in the compound name.
Beryllium: Be+2
Phosphate: PO4-3
The second thing you need to do is balance the charges, so they equal zero. The two charges we have at the moment are +2 and -3, which add up to -1, not 0. To make them equal zero, you need different amounts of the chemicals, which would multiply the charges.
Be+2 PO4-3
With this, we would have three beryllium ions and two phosphate ions.
2*3 = 6 -3*2= -6
Making this the formula name for beryllium phosphate:
Be3(PO4)2
See, the charges are nullified. If the parentheses are confusing, remember that they just keep the ion together, and the subscript two after the parentheses means that there are two phosphate ions.
_______________________________________________________
For the second, easier, example, I'll do potassium chlorate:
First, write down the ions in the compound name.
Potassium: K +1
Chlorate: ClO3-1
Again, you have to balance the charges so that they add up to 0.
K +1 ClO3-1
1+-1 = 0
These ions already balance each other out, so you don't have to do anything.
So, potassium chlorate is KClO3 .
________________________________________________________
\if that makes sense
Friday, February 8, 2013
1st
This is a blog for Chemistry class. On here will be... chemistry? I don't know what this is for really... so yeah.. for science!
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