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Homework Help: Heat transfer calculation problem

  1. Sep 27, 2014 #1
    I must be doing something wrong here but I cannot figure out what it is

    1. The problem statement, all variables and given/known data

    An insulated beaker with negligible mass contains a mass of 0.350kg of water at a temperature of 70.1∘C.
    How many kilograms of ice at a temperature of − 11.6∘C must be dropped in the water to make the final temperature of the system 25.0∘C?

    Take the specific heat for water to be 4190J/(kg⋅K) , the specific heat for ice to be 2100J/(kg⋅K) , and the heat of fusion for water to be 334kJ/kg .

    2. Relevant equations

    Q = mcT
    Q = mcT + Lfm

    3. The attempt at a solution

    So basically I was just doing...

    mwcw ΔTw = mi [(ci ΔTi) + Lf i]

    which gives
    ΔTw = 70.1∘C. - 25∘C = 45.1K
    ΔTi = − 11.6∘C - 25∘C = 36.6K

    0.350kg * 4190J/(kg⋅K) * (45.1K) = mi [(2100J/(kg⋅K)*(36.6K)] +334kJ/kg]
    66139.15J = 257140J/kg * mi
    mi = 0.161kg

    which does not turn out to be the right answer (I do not have access to the right answer).
    Can anyone tell me what mistake I am making.
    Any help would be greatly appreciated.
  2. jcsd
  3. Sep 27, 2014 #2


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    Staff Emeritus
    Science Advisor
    Homework Helper

    What happens to the ice when it reaches a temperature of 0 C? Is your equation equating the heat lost by the water in the beaker to the heat gained by the ice dropped into the beaker correct? You should first write out what happens to the ice after it is dropped into the beaker without applying any numbers.
  4. Sep 27, 2014 #3
    Q = mi [(ci ΔTi) + Lf i]

    Would this equation not mean that each kilo of ice absorbs the heat needed to change its state and also to change its temp?
    for each kilo 2100 Joules per degree change in temp but also the 334kJ to change its state?

    Am i right in assuming that each kilo of ice changes through -11.6∘C to 0∘C, taking 24360J , changing state for 334000J, then continuing to change from 0∘C to 25∘C taking a further 52500J. Total = 410860J/kg

    Whoops, that's different from above, but It is what I used in my calculations
  5. Sep 27, 2014 #4
    oh... It becomes water and has a different specific heat after 0∘C.... hmmmm. Ill try that
  6. Sep 27, 2014 #5
    That worked.

    Qwater = Qice + Qicewater
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