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Heat and temperature change in liquid

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. Portion A has a mass m and a temperature of 94.0 °C, portion B also has a mass m but a temperature of 78.0 °C, and portion C has a mass mC and a temperature of 34.0 °C. What must be the mass of portion C so that the final temperature Tf of the three-portion mixture is Tf = 50.0 °C? Express your answer in terms of m; for example, mc = 2.20 m.



    2. Relevant equations

    C= (m(Tf-T0))/ Q



    3. The attempt at a solution

    I attempted to fill in the equation for specific heat for each portion and set them equal to each other, because they are the same liquid/same value for C. This just started me going around in circles; I'm clearly going about this the wrong way.

    Could anyone provide some guidance to get me started on the right track? Thanks so much for your time.
     
  2. jcsd
  3. Dec 6, 2011 #2
    sorry, typo.

    for the equation I meant: Q = cm (Tf- T0)
     
  4. Dec 6, 2011 #3
    The heat lost by portion A plus the heat lost by portion B equals the heat gained by portion C if the final temperature is 50 deg C

    Write this down as an equation in terms of m, c and the change in temperature for each portion. Rearrange/solve for mC
     
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