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Gumby
Fishnologist


Registered: 06/13/01
Posts: 26,656
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Anyone know thermochemistry?
#5814076 - 07/02/06 11:07 PM (17 years, 6 months ago) |
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I figured someone here has done thermochem, so I'm going to ask for your help on a problem:
Here's the problem: A 500 g sample of iron at 1100ºC is dropped into a 20 quart vat of water. The initial water temperature is 20C. The specific heats of iron and water are 0.448 and 4.184 J/gºC, respectively. What is the final temperature in degrees Celsius of the mixture? The density of water is 1.00 g/cm3 at 20ºC.
Here's what I have so far: -(500g)(0.448J/gºC)= 244J/ºC <=== heat capacity(C) of iron -20 quarts= 19 liters = 19kg/19000g water -(19000g)(4.184J/gºC)= 79496J/ºC <== heat capacity(C) of water
Here's what I don't get. To calculate the heat change I use this equation: q = C(Tf-Ti)
For both water and iron, I have Ti, but not Tf. Even if I did have Tf, I wouldn't know what to do from there.
Can anyone offer any help?
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blacksabbathrulz

Registered: 05/22/02
Posts: 2,511
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Re: Anyone know thermochemistry? [Re: Gumby]
#5814231 - 07/03/06 12:00 AM (17 years, 6 months ago) |
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23.1 degrees I believe. I believe you substitute the heat supplied in the equation, use 19500 as the mass, ignore the Cp of iron at this point since it was factored into the heat supplied, and solve. I'm probably wrong though. If I had my thermal book handy I'd be more help.
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Gumby
Fishnologist


Registered: 06/13/01
Posts: 26,656
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Holy shit man, thats right. You own. Thanks a ton. I've been trying to figure this one out for an hour and a half.
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Gumby
Fishnologist


Registered: 06/13/01
Posts: 26,656
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I got so excited I didn't take the time to ask a question...
What do you mean you subsitute the heat supplied?
If you don't mind, will you show me how you set the equation up to give you 23.1C?
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blacksabbathrulz

Registered: 05/22/02
Posts: 2,511
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Re: Anyone know thermochemistry? [Re: Gumby]
#5814292 - 07/03/06 12:28 AM (17 years, 6 months ago) |
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Sure thing, just give me a minute here and I'll post how I did that.
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blacksabbathrulz

Registered: 05/22/02
Posts: 2,511
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Okay, so I multiplied 500 grams, which was the weight of the iron times 1100 degrees C, and then by 0.448 J/g degree c. The grams and degrees c cancel out leaving you with 246,400 joules being supplied. Substitute that for Q. You know that the weight of the water is 19000 grams, and you are adding 500 grams, so your total weight is 19500 grams. Divide the 246400 joules by the weight and the Cp of water to get the rise in temperature. Add this to 20 degrees C and you have your final temp. I'm fairly certain something about this is slightly flawed, however I think it should get you the approximate temp. Do you happen to know if my answer is correct? You said that it was right, but I wasn't sure if you meant the rationale was right, or the answer. Hope I helped.
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Gumby
Fishnologist


Registered: 06/13/01
Posts: 26,656
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Yep, your answer was correct. Thanks a bunch. Knowing what you told me is going to help me a lot because I have a lot of "Find the final temperature" type problems.
Thanks again man
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blacksabbathrulz

Registered: 05/22/02
Posts: 2,511
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Re: Anyone know thermochemistry? [Re: Gumby]
#5814454 - 07/03/06 01:27 AM (17 years, 6 months ago) |
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Quote:
Gumby said: Yep, your answer was correct. Thanks a bunch. Knowing what you told me is going to help me a lot because I have a lot of "Find the final temperature" type problems.
Thanks again man
No prob. If you have any other questions let me know, and I'll see if I can help. It's been a while since I did this stuff, but I'm glad to see I did alright.
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