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Homework Help: Calorimetry with ice and water

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data
    A block of ice at 0C is added to a 150g aluminum calorimeter cup which holds 210g of water at 12C. If all but 2.0 g of ice melt, what was the original mass of the block of ice

    2. Relevant equations

    heat gained by ice = heat lost by water + heat lost by calorimeter

    3. The attempt at a solution
    so i figure that its:
    (m(i)+2.0x10^-3kg)*c(i)*deltaT(i) = (m(w)*c(w)*deltaT(w) +(m(a)*c(a)*deltaT(a), where m is mass, c is the specific heat and T is temperature
    but what i dont get is isnt the deltaT(i) going to be zero since the ice starts off zero and the water/ice mixture temperature zero? so then this equation doesnt work because solving for m(i), you cant divide a number by zero?
  2. jcsd
  3. Apr 12, 2010 #2
    You need to use the latent heat of fusion to find out how much heat it takes to melt ice.
  4. Apr 12, 2010 #3
    so instead of m(i)*c(i)*deltaT(i) in the above equation, I use m(i)*L, where L is 333 kJ/kg?
  5. Apr 13, 2010 #4
    Yeah, it's a phase change.
  6. Apr 13, 2010 #5
    so i did what you said and this is what i got
    (m(i)+2.0x10^-3kg)L = (m(w)*c(w)*deltaT(w) +(m(a)*c(a)*deltaT(a)
    (m(i)+2.0x10^-3)(333) = (210x10^-3*4186*(12-0)) + (150x10^-3*900*(12-0))
    m(i) = 36.54kg
    this doesnt make sense does it??
  7. Apr 13, 2010 #6
    Wait, shouldn't your specific heats stay at the same numbers?

    4.186 J/goC = 4.186 kJ/kgoC
  8. Apr 14, 2010 #7
    what do you mean? little confused ... you said to use the m(i)*L for the left side of the equation and then for the right side, don't i use the specific heats for water and aluminum?
  9. Apr 14, 2010 #8
    You do, but you messed up the numbers. You used 4186 J/kgoC instead of 4.186 kJ/kgoC. So your answer comes out off by a factor of 1000.

    Make sure you check that your units are consistent before you answer the question (you used 333 kJ/kg for latent heat so every value should involve kJ and/or kg).

    Also, this term is incorrect. (m(i)+2.0x10^-3)(333) All the initial mass BUT 2 g melts so you should subtract.
  10. Apr 14, 2010 #9
    alright .. i get it now .. thanks for the help!
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