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I really have no clue how to solve this.
"Given 360.5g of hot tea at 80 (C), what mass of ice at 0 (C) must be added to obtain iced tea at 10.5 (C)? The specific heat of the tea is 4.18, and the heat of fusion for ice is 6.01 kj/mol.
 one year ago
 one year ago
I really have no clue how to solve this. "Given 360.5g of hot tea at 80 (C), what mass of ice at 0 (C) must be added to obtain iced tea at 10.5 (C)? The specific heat of the tea is 4.18, and the heat of fusion for ice is 6.01 kj/mol.
 one year ago
 one year ago

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augusto.hBest ResponseYou've already chosen the best response.1
(tea) mass*spec heat*T = (water) m*heat fusion
 one year ago

SchrodingerBest ResponseYou've already chosen the best response.0
Could you explain in words how you would arrive at the point/how i'm supposed to know that association? Thanks.
 one year ago

augusto.hBest ResponseYou've already chosen the best response.1
Actually I think there is something wrong because tea doesn't freeze at 10,5ºC  1 atm. Unless the pressure is other. If so you will need the heat of fusion of tea as well.
 one year ago

augusto.hBest ResponseYou've already chosen the best response.1
Are you sure about the question?
 one year ago

SchrodingerBest ResponseYou've already chosen the best response.0
Let me double check: I'll directly copy/paste it to make sure there aren't any possible errors on my part: "Given 360.5 of hot tea at 80.0, what mass of ice at 0 must be added to obtain iced tea at 10.5? The specific heat of the tea is 4.18 , and for ice is +6.01 ."
 one year ago

augusto.hBest ResponseYou've already chosen the best response.1
Well I don't know. It seems to me there is lack of info. I would do this. Calculate the hole energy for the tea process. Q total = Q1 (T80>Tfusion) + Q2 (fusion) + Q3 (Tfusion>T10.5 if 10.5 is not the fusion T o f tea). Then you use this energy to calculate the mass of water Q = m*L where L is 6.01kj/mol
 one year ago
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