Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: mr cool on February 08, 2009, 11:29:22 AM
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I did a lab on friday and we made suckers with corn syrup and sugar.
This is the question I am stuck on...
You know the rate (can be calculated) of energy input into your system and the mass of the corn syrup and sugar mixture. What is the specific heat of this sugar solution
first of all: I dont know how to calculate the rate, and i dont know the mass of the mixture. I know the density of the mixture was 1.30g/ml
we used 20.00 ml of water
30.00 g of sugar
and 20.00ml of corn syrup
to get the mass could i divide 20.00 ml of corn syrup by 1.30 g ml (density) and come out with mass?
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edit: it wouldn't let me modify so i had to re-post...
alright the mass of my solution is 45.38 g.
this is the formula for specific heat... Q = m*c* deltaT
Q = Specific heat
M = mass
C = The quantity of energy needed to raise the temperature of one unit of material 1 degree.
deltaT = change in temperature
i have trouble with C because i have no clue how much energy it takes to raise corn syrup + sugar one degree.
Can someone help me out with this it really sucks
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I imagine you somehow measured the temperature change, and also how much energy you put into the system? Can you post your procedure?
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I imagine you somehow measured the temperature change, and also how much energy you put into the system? Can you post your procedure?
yes but it is a long one... (oh yes we did keep track of temp via thermometer)
(Irrelevant) 1: You will not us any glassware from you lab drawer during this lab.
(Irrelevant) 2: Make an aluminum ring mold for your sucker, and tape it securely on wax paper. Insert a sucker stick through the aluminum mold.
3: Record the mass of 30.00g of sugar and add to a 600-ml beaker.
4: Record the volume of 20.00 ml of water and add to a 600-ml beaker.
5: Record the volume of 20.00 ml of corn syrup and add to a 600-ml beaker. Record the total mass of the solution. Density of corn syrup is 1.30 g/ml.
6: Heat contents of the 600-ml beaker over a Bunsen burner with constant stirring.
7: Record the temperature of the solution every 30 seconds
8: Remove the beaker from the heat when the temperature reaches 148 - 154 oC
(Irrelevant) 9: Continue to stir as the solution begins to cool and thicken. Do not allow the solution to harden in the beaker.
the rest of the steps involve us pouring our solutions into the molds.
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Hmm...I'm pretty sure that you need the amount of energy that was input, not just the temperature change in order to calculate the specific heat. I'm not sure how you would go about getting that, sorry.
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Hmm...I'm pretty sure that you need the amount of energy that was input, not just the temperature change in order to calculate the specific heat. I'm not sure how you would go about getting that, sorry.
The last question on the sheet is...
You know the rate (can be calculated) of energy input into your system and the mass of the corn syrup and sugar mixture. What is the specific heat of this sugar solution
so i guess i have enough information to calculate it?
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Hm...I guess I'm just not seeing the relation. It'll probably be an "aha" moment when someone does give us a thought how to do it.
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Hm...I guess I'm just not seeing the relation. It'll probably be an "aha" moment when someone does give us a thought how to do it.
to find energy of something (worded badly) you just have to do Calories (ie 1 cal/g) * Mass (45.38g) * DeltaT (148oC - 35.1oC) = 5123 calories? i think that is how much energy was used if im correct?
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Thanks ARGOS for verifying my thoughts. I agree, we need the amount of heat used.
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Thanks ARGOS for verifying my thoughts. I agree, we need the amount of heat used.
you cant figure that out by using the temperatures?
the main objective in this lab is to find the specific heat of the solution. i dont think he would set us up for failure.
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On my graph there is a plateau around 100*C . And then goes up even faster than before. Why does the temp raise quicker than before the plateau?
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But the slope would be dependent upon the amount of energy that was being output, and that is completely dependent on the source, which could be anything. I really don't think this question is possible without giving the amount of energy consumed by the system.
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Ok here we go....
Time | temp
.5. 35.1
1. 57.1
1.5. 63
2. 65.1
2.5. 89
3. 91
3.5. 95
4. 101
4.5. 104
5. 104
5.5. 104
6. 104
6.5. 104
7. 104
7.5. 106
8. 106
8.5. 107
9. 108
9.5. 110
10. 113
10.5. 115
11. 117
11.5. 120
12. 130
12.5. 135
13. 142
13.5. 145
14. 148
Some stuff from the graph....
Y = 5.7563x + 62.563
R2 = 0.8402
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Thanks anyways. I called some of my chem friends and they had no clue how to do it either so i will probably go into school early tomorrow.
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Here is how we did it...
Take the heat of vaporization X mass of H20
(540 cal/g)(20.00)g = 18900 cal
Divide that by the time it took to vaporize all the water
10800cal/g 3.6 min = 3000 calories
The ammount if energy doesnt change so 18900/ mass / deltaT
(i forgot where i got 18900 from but it is how much energy was used in the final portion of the graph.)
It came out to around 7 something just for those who were wondering about this yesterday. So i guess my chem teacher > yours :p
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Dear reezyfbaby;
Thank you for all that information!
So it is basically the method I explained in my former reply (see: #15) to get from the plateau the heat transfer of the Bunsen burner and from the slope the amount.
Even as it is a very rudimentary and very imprecise method, it is worth to be remembered.
Good Luck!
ARGOS++