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Schrodinger
 4 years 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.
Schrodinger
 4 years 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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(tea) mass*spec heat*T = (water) m*heat fusion

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are you sure about the question?

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0Let 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 ."

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well 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
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