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Homework Help: Specific latent heat and heat capacity

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    0.1kg of ice was added to a lemonade of mass 330g at an initial temperature of 28 degree celcius and it completely melted. Considering only the lemonade and ice cubes, calculate the final temperature of the lemonade.

    Given: specific latent heat of fusion of ice= 336000 J/kg
    specific heat capacity of lemonade= 4200 J/kg


    2. Relevant equations



    3. The attempt at a solution

    total energy loss by ice cubes= 0.1kg * 336,000J
    =33,600J

    heat capacity of 330g of lemonade=0.33kg * 4200 J/kg
    = 1,386J

    Total temperature fall by lemonade=33,600J / 1,386J
    =24.24242424

    Final temperature=Initial temperature - fall in temperature
    = 28-24.242424
    = 3.80 degree celcius (3 s.f.)
     
  2. jcsd
  3. Apr 24, 2010 #2
    Bring up my post
     
  4. Apr 25, 2010 #3
    Help will be appreciated...
     
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