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Solar-heated water, farmer

  1. Nov 4, 2014 #1

    rlc

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    1. The problem statement, all variables and given/known data
    To help keep his barn warm on cold days, a farmer stores 859 kg of solar-heated water (Lf = 3.35E+5 J/kg) in barrels. For how many hours would a 2.02 kW electric space heater have to operate to provide the same amount of heat as the water does, when it cools from 11.3 to 0 °C and completely freezes?

    2. Relevant equations
    Q=mc(deltaTdegC)+mL

    3. The attempt at a solution
    Q=(859)(4187)(11.3)+(859)(335000)
    Q=328406952 J
    1kW=1000W=1000J/sec
    1000t=328406952 J
    t=328406 secs*1min/60secs*1hour/60mins=91 hours

    But this is wrong. What am I missing?
     
  2. jcsd
  3. Nov 4, 2014 #2

    NascentOxygen

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    Staff: Mentor

    You have calculated for a 1000w heater.
     
  4. Nov 4, 2014 #3

    collinsmark

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    Homework Helper
    Gold Member

    You're correct that 1 kW = 1000 J/sec. But the heater is a 2.02 kW heater, not a 1 kW heater.

    [Edit: like NascentOxygen said.]
     
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