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How much energy is transferred to the water as heat?

  • Thread starter VitaX
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  • #1
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Homework Statement



A 140 g copper bowl contains 190 g of water, both at 21.0°C. A very hot 450 g copper cylinder is dropped into the water, causing the water to boil, with 11.8 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy (in calories) is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature (in Celsius) of the cylinder? The specific heat of water is 1 cal/g·K, and of copper is 0.0923 cal/g·K. The latent heat of vaporization of water is 539 Cal/kg.

Homework Equations



Q = cm(Tf - Ti)
Q = Lm

The Attempt at a Solution



Qw = (1 cal/g*C)(190g)(100 C - 21 C) = 15,010 cal
Qb = (.0923 cal/g*C)(140g)(100 C - 21 C) = 1020.8 cal
Qc = (.0923 cal/g*C)(450g)(100 C - Ti) = 4153.5 - 41.5
Lm = (539 cal/g)(11.8g) = 6360.2 cal

Qc = Qw + Qb + Lm
4153.5 - 41.5Ti = 15,010 + 1020.8 + 6360.2

Ti = -439.46 C

The energy transferred in this problem is just Q right? So the energy transferred to the bowl would be what Qb is equal to? My final answer for the cylinders initial temperature is clearly wrong. What did I screw up in my work? Isn't it Q lost = Q gained?
 
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Answers and Replies

  • #2
gneill
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I think you'll find that the heat of vaporization of water that you're using is about 1000X too small. Evidently the capital "C" on Cal was meant to indicate kcals.
 
  • #3
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I think you'll find that the heat of vaporization of water that you're using is about 1000X too small. Evidently the capital "C" on Cal was meant to indicate kcals.
You mean where it says Lv = 539 Cal/Kg that it is actually Kilocalories? So it should be 539000000 cal/g? Converted Kg to g and Cal to cal. Something just does not seem right here. Either way my answer is going to be negative with my current equation. What is the correct equation for finding the initial temperature?

Is my equation right for heat loss = heat gain? Another question is, when do you set up all the Q's gathered in a question equal to 0? Is that only when the problem states it's an isolated system?
 
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  • #4
gneill
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You mean where it says Lv = 539 Cal/Kg that it is actually Kilocalories? So it should be 539000000 cal/g? Converted Kg to g and Cal to cal. Something just does not seem right here. Either way my answer is going to be negative with my current equation. What is the correct equation for finding the initial temperature?

Is my equation right for heat loss = heat gain? Another question is, when do you set up all the Q's gathered in a question equal to 0? Is that only when the problem states it's an isolated system?
The heat of vaporization for water is 2257 kJ/kg, or 539000 cal/kg, or 539 cal/g. I was thrown by the "539 cal/kg" figure.

The only problem I see with your equations is that for your calculation of Qc, the ∆T should be going in the other direction -- the Ti will be higher than 100C. So either reverse their order, or use -Qc subsequently.
 
  • #5
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The heat of vaporization for water is 2257 kJ/kg, or 539000 cal/kg, or 539 cal/g. I was thrown by the "539 cal/kg" figure.

The only problem I see with your equations is that for your calculation of Qc, the ∆T should be going in the other direction -- the Ti will be higher than 100C. So either reverse their order, or use -Qc subsequently.
Ah I was wondering about that initially. But I went with 100 C as the final temp because the way it was worded in the problem. So basically with whatever loses heat, I should always put a negative sign in front of that Q if I'm going to have a higher initial temp than final temp for it? Does the energy transfer just equal Q for that certain substance as well? And I can just use 539 cal/g for this problem right?
 
  • #6
gneill
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Ah I was wondering about that initially. But I went with 100 C as the final temp because the way it was worded in the problem. So basically with whatever loses heat, I should always put a negative sign in front of that Q if I'm going to have a higher initial temp than final temp for it? Does the energy transfer just equal Q for that certain substance as well? And I can just use 539 cal/g for this problem right?
The best way to keep track of the signs of things is to know what's happening (like the cylinder losing heat, not gaining it), and knowing how you're going to be using the value later on in your calculations. It's perfectly okay to calculate the magnitudes of things along the way (so that all the values are positive) so long as when you put them together they take on the appropriate sign in the operations.

Yes, 539 cal/g is okay.
 
  • #7
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I just input my answers into the homework online and it said I got all parts right but part A. I don't see what I did wrong because using what I got as Q for water I had to find the initial temperature of the cylinder which I got right. What should the energy transfer for the water be here if not Q for the water?
 
  • #8
gneill
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I suppose one could argue that the heat transferred to the water includes the heat of vaporization to turn part of the water to steam. Steam is still water, after all.
 
  • #9
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I suppose one could argue that the heat transferred to the water includes the heat of vaporization to turn part of the water to steam. Steam is still water, after all.
Ahh. I'm betting you are right.
 

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