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Melting ice

  1. Dec 11, 2005 #1
    How much ice could be melted by 320. grams of water at 68.0 degrees C?
    Would I us mL + mcT=0 for the equation, and do the latent heat of ice for the first part and water for the second?
     
  2. jcsd
  3. Dec 11, 2005 #2

    Astronuc

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    Staff: Mentor

    One must determine the energy (heat) from the mass of water decreasing from 68°C to 0°C. Then determine what mass of ice would be transformed to water at 0°C.

    Assuming that L in your equation is latent heat, and by T you mean change in temperature, you are correct.

    The heat from the liquid, Q, is just the change in energy in the liquid, i.e. Q = [itex]\Delta\,H[/itex] = [itex]m\,c\,\Delta\,T[/itex], where c is specific heat.
     
  4. Dec 11, 2005 #3
    So it would be m(333.7kJ/kg)+(.320kg)(4.186kJ/kg*K)(-68.0K)=.273kg of ice Is this correct?
     
  5. Dec 11, 2005 #4

    Astronuc

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    Staff: Mentor

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