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Heat and phase change

  1. Jan 24, 2007 #1

    1. A 10-kg block of ice has a temperature of -10°C. The pressure is one atmosphere. The block absorbs 4.12E6J of heat. What is the final temperature of the liquid water?



    2. Q = mL and Q = cm/\T



    3. I tried doing Q = cmT and plugging in the values above to get 88.4 degrees as the final temperature, and using 4186J/(kgCdegree) for c, but it's wrong. I am confused on these types of problems.

     
  2. jcsd
  3. Jan 24, 2007 #2
    you've ignored the latent heat required for the phase change from ice to liquid water...
     
  4. Jan 24, 2007 #3
    I know, but I don't understand how it fits in.

    mL + cmT = cmT?
     
  5. Jan 24, 2007 #4
    you could try working out how much specific heat would be needed to raise it to melting point, then the latent heat needed to melt it, then subtract both these values from the total energy, and do one more Q=cmT calculation to work out the final temperature.
     
  6. Jan 24, 2007 #5
    what is the total energy, and how would that work, subtracting Q-L if they both have different units?
     
  7. Jan 24, 2007 #6
    sorry, bit ambiguous there. By total energy, I meant the number you were given for the total work applied.

    I would approach this problem like this (there's probably a more elegant way, and if you want to leave the terms algebraic and cancel down etc, that might help. Given a calculator, many people don't find it necessary)

    1. use Q=mcT to calculate how much heat one would have to apply to the water to heat it to 0 degrees (i.e T = 10). Call this Q_1 for example

    2. use Q = mL to calculate the latent heat that will be required to change the material to a liquid. Call this amount of heat Q_2 perhaps.

    3. now, Q_1 + Q_2 is the total heat you have "used" so far. Subtract this from the amount of heat that was applied in total: 4.12E6 - Q_1 - Q_2. Call the remaining amount of heat Q_3

    4. finally, work out the temperature change you get from applying this remaining amount of heat Q_3 to the (now liquid) water, starting at 0 degrees.
     
  8. Jan 24, 2007 #7
    thank you very much, you helped me a lot.
     
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