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Homework Help: Thermal physics problem

  1. Apr 21, 2005 #1
    Please help.

    An insulated beaker with negligible mass contains liquid water with a mass of 0.350kg and a temperature of 76.3 degrees celsius.

    How much ice at a temperature of −18.1 degrees celsius must be dropped into the water so that the final temperature of the system will be 35.5 degrees celsius?

    Take the specific heat for liquid water to be 4190 J/kg.K, the specific heat for ice to be 2100 J/kg.K}, and the heat of fusion for water to be 334kJ/kg.

    My final answer was 0.12kg, , but the feedback i received was "you are close". I can't seem to get the right answer for it. Please help.

  2. jcsd
  3. Apr 21, 2005 #2
    Please help.
  4. Apr 21, 2005 #3
    If you were close to the answer, explain us the steps of your reasoning.
  5. Apr 21, 2005 #4
    okie dokie... i did m_ice*c_ice*(0--18.1)+ m_ice*L+ m_ice*c_ice*(35.5-0) *m_water*c_water*(35.5-76.3).

    I think my mistake may have been with the heat capacity of the ice. I think i need to put 4190 instead of 2100 when the ice is melted.
  6. Apr 21, 2005 #5
    Yep, you're close to the answer, except two things. First, substract the lower temperature from the higher temperature. Second, your mistake, after the ice has melted don't use c(ice) any more, but c(water).
  7. Apr 21, 2005 #6
    But shouldn't the change in temp be t_final-t_initial?
  8. Apr 21, 2005 #7
    Ok, it depends on the context where you use it. I was thinking that you were using the equality [tex]Q_{gained} = Q_{lost}[/tex], meaning that the heat gained by ice is equal to the heat lost by water. T_final - T_initial would make the RHS negative.
    On the other hand, if you use [tex]Q_{gained} + Q_{lost} = 0[/tex] it is ok to use as you did, assuming that the lost energy is negative.
  9. Apr 21, 2005 #8
    If you are referring to water, i think it is menat to be a negative number, because I'm using the mastering physics thing, and it says it's correct.

    Thanks for your help.
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