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Thermodynamic process: freezing

  1. Nov 3, 2016 #1
    1. The problem statement, all variables and given/known data

    A 0.350 kg sample is placed in a cooling apparatus that removes energy as heat at a constant rate. The figure below gives the temperature T of the sample versus time t; the horizontal scale is set by Ts = 80.0 min. The sample freezes during the energy removal. The specific heat of the sample in its initial liquid phase is 3260J/kg·K. (a)What are the sample’s heat of fusion and (b) its specific heat in the frozen phase?

    upload_2016-11-3_8-0-41.png
    2. Relevant equations
    Q=mL
    Q=mcdelta T
    P=Q/t
    3. The attempt at a solution:
    constat rate: Q=c.m.delta T=3260x0.35x30=34230 J
    P=q/t=34230/(40x60)=14.2625 W
    a) q=ml (negative since itis a solidification)
    but q=pt=14.2625x30x60=2567.2=mL====> L=2567.2/0.35=73350 J/Kg = 73.35 KJ/Kg
    b) q=cxmxdeltaT but Q=pt=14.2625x20x60=17115J ====> c=17115/(mx delta T)= 17115/(0.35x20)=2445 J/Kg.K
    I want only to check myanswerif correct
     
    Last edited: Nov 3, 2016
  2. jcsd
  3. Nov 3, 2016 #2

    DrClaude

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    Staff: Mentor

    What is the question you have to answer?
     
  4. Nov 3, 2016 #3
    What are (a) the sample’s heat of fusion and (b) its specific heat in the frozen phase?
    Sorry, I thought I copied it with the question
     
  5. Nov 3, 2016 #4

    DrClaude

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    Staff: Mentor

    Looks ok. (Although I do not like the lack of units in the intermediate steps of the calculations :smile:)
     
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