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Thermodynamics: Q=mc(deltaT) 
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#1
Nov1608, 12:51 PM

P: 9

1. The problem statement, all variables and given/known data
In a physics lab experiment a student immersed 191 onecent coins (each having a mass of 3.00 g) in boiling water, at a temperature of 100 degrees C. After they reached thermal equilibrium, she fished them out and dropped them into an amount of water of mass 0.275 kg at a temperature of 17.0 degrees C in an insulated container of negligible mass. What was the final temperature of the coins? (Onecent coins are made of a metal alloy  mostly zinc  with a specific heat capacity of 390 J/(kg * K).) Use 4190 J/(kg* K) for the heat capacity of the water. 2. Relevant equations Q=mc(deltaT) Q1 + Q2=0 3. The attempt at a solution I'm not sure where you can get a workable system of equations from this problem. I tried finding the Q of the coins, which I found to be Q=(191*.003kg)(390)(T100) where T=the final temperature of the coins. And the Q of Water: Q=(.275)(4190)(T17). You also know that the Q of water and the Q of the coins must be equal to 0 because they are in an insulated container. Because the water is heated and the coins are cooled, you know that the Q for water is positive and that Qwater=Qcoins, However, they have different final temperatures and I'm not sure how to find one in terms of the other so I can solve the equation I've come up with. Please help! 


#2
Nov1608, 01:08 PM

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P: 4,300

I don't know if you did it on purpose or by accident, but you used the same letter T in both equations. Is it really the same? If not, use a different letter. But if you can argue they are, you can set Qwater = Qcoins which gives you: 191*0.003*390*(T  100) = 0.275*4190*(T  100) and solve for T (one equation, one unknown).



#3
Nov1608, 01:35 PM

P: 9

You can't assume the final temperatures are the same, sorry. So Qcoins=Qwater but
(191*.003)(390)(R100)=.275(4190)(T17) 


#4
Nov1708, 09:08 AM

P: 9

Thermodynamics: Q=mc(deltaT)
I guess you can assume the final temperatures are equal. I think it was a problem with my signs in that I overcompensated for the fact that the coins lose heat, so they have a negative value for Q. The final temperature will obviously be less than 100 so I suppose that accounts for the sign change.



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