Today, we started off class my going over the last two questions on last night's worksheet, which you can find...here. The answers are here! I shall explain the last two for any of you who are still wondering...
8. The key to this problem is that the heat lost by the copper is equal to the heat lost to the water. In the problem, they give you all three variables you would need to find the heat lost (the mass, the specific heat capacity, and the temperature change). However, to find the heat lost by the water, you still need to figure out the mass of the water. So, you set the two equations equal to each other, remembering that the heat lost by both of them is the same.
(110 g)(.20 J/g°C)(57.5°)=(m)(4.18 J/g°C)(2.6°C)
m= 116.4 g H20
9. Finding the specific heat capacity is simple in this question. You just must use the knowledge they give you to solve for that variable.
585 J=(125.6 g)(23.5°C)(shc)
shc=.14 g/°C
Then, using that answer, you just convert it to moles using the molar mass!
(.14 g)(200.59 g)=30.67 J/mol
Next, we moved onto the lab called the Specific Heat of Metal Lab. Mr. Liebs (follow him on Twitter-...@davejlieberman)checked in all our pre-labs while we set up the lab that was to be worked on with our partners (mine being Michelle). Get the lab here if you missed it.
For the pre-lab, number one asked you to calculate the energy transfer.
1. (300 g)(4.18 J/g°C)(37°C)=46,398 J.
Number two asked why you shouldn't eat snow if you're trapped in Alaska in place of water. That is because your body uses up so much energy trying to melt the snow that it dehydrates you. All of your body's energy is being transferred to the snow! So just remember, melt the snow before you attempt to consume it.
First, we turned up the hot plate all the way so that we could go the water in the 600 mL beaker to a boil.
Then, we got a dry test tube and measured the weight, which was 12.01 g. We took a random container from the counter, which was labeled with a "B" and poured all the contents into the test tube.
We then measured the total amount, which was 60.0 g.
While we were waiting for the water on the hot plate to boil, we got a calorimeter, which strongly resembled a styrofoam cup, from the counter and poured 51 mL of water into it. The temperature of the water was 22.6°C. We placed the top over the calorimeter to insulate the water.
While we waited for the water to boil, I snapped a couple of pictures of some of my peers, which they promptly made me delete. I thought they were good! But, whatever...
When the water finally came to boil, Michelle and I, along with the other duo at our table, Ben W. and Aaron, placed our test tubes filled with mystery metal into the beaker.
After waiting, 10 minutes, we took the beakers out carefully, turned the hot plate off and spilled the metal into the calorimeter. While it was still hot, we recorded the new temperature of the water inside the calorimeter. It came out to be 30.6°C.
We then cleaned up our lab tables and head back to our desks to try out the post-lab.
I don't know why there's this weird space below, but I don't know how to fix it, so...this is my data table!
My Data Table:
Mass of test tube | 12.01 g |
Mass of test tube & metal | 60.0 g |
Label on container | B |
Volume of water | 51 mL |
Initial temp of water | 22.6°C |
Initial temp of boiling water | 100°C |
Final temp in calorimeter | 30.6°C |
These answers are according to my data, but I hope it helps!
1. The heat gained by the calorimeter
30.6°C-22.6°= 8°C
2. The heat lost by the metal
100°C-30.6°C= 69.4°C
3. The specific heat of the metal and what it is- for this question, you must set the heat lost by the metal to the heat gained by the calorimeter because you do not have the specific heat for the metal.
(51 g)(8°C)(4.18 J/g°C)=(47.99 g)(69.4°C)(specific heat)
specific heat= .512 J/g°C
This number is not exactly close to any metal on the worksheet, but it is closest to zinc and copper, and since the metal looked more silvery and not reddish, I assume it was zinc.
4. Percent error. Which is the absolute value of actual-theoretical divided by actual times 100 if anyone forgot.
|.385-.512|/.385*100= 33%
Yikes! This is obviously not a very good percent error, but there were things in the lab that could've been done slightly wrong. We may have not kept the thermometer in the water for long enough, or maybe we did a bad massing job. Some of our measurements could've been off, but in the end, we were able to figure out our mystery metal.
That's about it for now! I hope you guys had a great break and happy 2011!
The next scribe is...my fabulous partner Michelle T.! Have fun :)
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