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

  1. Dec 9, 2009 #1
    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?
     
  2. jcsd
  3. Dec 9, 2009 #2
    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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