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How long to melt the ice(heat, heat transfer, latent heat)

  1. Jun 1, 2012 #1
    The graph is in the attachment!!! ( dont have to download to view the file)





    1. The problem statement, all variables and given/known data
    49a) A 0.25 kg piece of ice is warmed by an electric heater and the following graph of temperature is produced. Assume that there has been no loss of energy to surroundings.

    How much additional time after 150s will be required to melt all of the ice, assuming the power on the heater is constant?


    2. Relevant equations

    Q=m1c1Δt (c is specific heat capacity and Δt is change in temperature)

    Q=mLf

    3. The attempt at a solution


    Another way to word the question 49a would be to ask ‘ How long does it take for ice to get from -10 degrees to 0 degrees

    Q=mLf is for the amount of energy it takes to melt the ice.


    Q=m1c1Δt is for the amount of energy it takes to heat the ice.


    Q=m1c1Δt

    =(0.25)(2.1 X 103)(t2-t1)

    =(0.25)(2.1 X 103)(0-(-30) <------the ice is initially at -30 as you can see on the graph.(graph is in attachment, and can be viewed without downloading)



    =(0.25)(2.1 X 103)(0+30)

    Q=15750 J

    This is the amount of energy required to heat the ice.



    P = W/Δt (Δt is now for time)


    since W = ΔE = Q


    P= 15750 J/150s (this 150s was given, as you can see on graph)

    P = 105 W ( this remains constant as the question states)


    Q=mLf is for the amount of energy it takes to melt the ice.

    =(0.25)(3.3 X 105)

    Q = 82500 J (it takes this much energy to melt the ice)



    P=Q/Δt


    Q/P = Δt

    82500/105 = Δt


    785.7 s =Δt


    The question asks how much additional time it will take AFTER 150s, so we subtract 150 from 785.7s

    785.7-150
    =635.7s


    It will take an additional 635.7s to melt the ice.

    or 635.7/60

    = 10.5 mins or it will take 10.5 mins to melt the ice.


    Does this look right? If not can you show me how to get the correct solution? Or better yet, if i have the incorrect solution, get the right solution and show it to me and i will figure out on my own on how to get the right solution.

    Thanks :)

    (graph is in the attachment, as always you dont have to download to view file :) )
     

    Attached Files:

  2. jcsd
  3. Jun 1, 2012 #2

    gneill

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    Staff: Mentor

    For the time interval shown on the graph (150s), the ice temperature change is not 30C. It starts at -30C and rises to -10C over the 150s. This will change your heater power calculation and subsequent values.
     
  4. Jun 1, 2012 #3
    ahh... i see

    So...everything else looks good? the procedure and the steps i have and the method i used to get the solution is fine?


    Also the reason i put it to 0 degrees instead of -10 is because i always understood t2 was usually the melting point
     
  5. Jun 1, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Your methodology looks okay; you seem to understand the required stages for solution.
    A variable is what you say it is :smile: When analyses become more complex than "generic examples" you have to be flexible in your variable assignments :smile:
     
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