1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook