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Mass of ice to warm a liquid - do I have it set up correctly?

by Linus Pauling
Tags: correctly, liquid, mass, warm
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Linus Pauling
#1
Dec9-09, 10:30 AM
P: 190
1. An insulated beaker with negligible mass contains liquid water with a mass of 0.250 kg and a temperature of 76.1C.

How much ice at a temperature of -12.5C must be dropped into the water so that the final temperature of the system will be 33.0C? Take the specific heat of liquid water to be 4190 J/kg*K, the specific heat of ice to be 2100 J/kg*K, and the heat of fusion for water to be 334 kJ/kg.




2. Q = Mc deltaT
Q = +/- ML_f




3. warming the ice = Q1 = m_ice * c_ice * deltaT = m_ice * (2100) * (0 - 260.5K)

melting ice = Q2 = m_ice * L_f = m_ice * (334000 J/kg)

bring liquids to same temp. = Q3 = m_ice * c_water * deltaT = m_ice * 4190 * (306K - 0K)

cool water = Q4 = m_water * c_water * deltaT = 0.250 * 4190 * (306K - 349.1K)

Then, I did Q4 = Q1 + Q2 + Q3, and solved for m_ice:

-45147.25 = m_ice * (1282140 + 334000 - 547050)

m_ice = -0.0422, which I just took to be positive. I think I am at least on the right track, but have I set it up wrong since I get a negative number for mass?
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Linus Pauling
#2
Dec9-09, 10:43 AM
P: 190
Actually, I have to correct myself: I set the Q of cooling the liquid to final temperature equal to the other three (warming ice + melting ice + warming the new mass of liquid). Solving, I obtain the mass of the original ice to be 0.0906 kg.

Got it! Nevermind.


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