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Thermodynamics - First law Q

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi all, please see uploaded pics. I can do the first 3 questions but can't get my head around the 4th. Any help is appreciated.

    Capture_zpsiwkqpcld.jpg

    DSC_0157_zpswmfhsrg7.jpg
     
  2. jcsd
  3. Apr 8, 2015 #2

    Delta²

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    The rate at which heat is lost by the air [itex]P_{air}[/itex], equals to minus the rate [itex]P_{water}[/itex]at which heat is absorbed by the water. So it will be [itex]P_{air}=-P_{water}[/itex].

    [itex]P_{air}=C_{air}\dot{m}_{air}(47-167)[/itex], [itex]P_{water}=C_{water}\dot{m}_{water}(40-35)[/itex]
     
  4. Apr 8, 2015 #3
    Hi Delta,

    What have you denoted as C?
     
  5. Apr 8, 2015 #4

    Delta²

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    Its the specific heat capacity. For water is 4.17 and for air around 1.01 at the temperatures of this problem.
     
  6. Apr 8, 2015 #5
    I thought it was the specific heat capacity but I can only find the Cv/Cp for water in any of the tables from my text which was confusing me a bit.

    Cheers for the help!
     
  7. Apr 8, 2015 #6
    It's liquid water. What is the heat capacity of liquid water?

    Chet
     
  8. Apr 8, 2015 #7
    Sorry in my previous post i meant to say CANT find the Cv/Cp value for water in any of my tables.
     
  9. Apr 8, 2015 #8
    The heat capacity of liquid water is 4.18 kJ/kg C. Did you not see Δ2's post # 4? Are you saying you never had this in freshman physics?

    Chet
     
  10. Apr 8, 2015 #9
    Yes I did see that in Delta's post. I am just saying, naturally, I refer to the tables for my data values and when I didn't come across a value for water it threw me off. I thought I had done something wrong. That's what happens when you have been studying all day I guess, time to take a rest.
     
  11. Apr 8, 2015 #10
    All I did was Google "heat capacity of water."

    Chet
     
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