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Compute Heat capacity

  1. Jan 3, 2012 #1
    You warm 1.00 kg of water at a constant volume of 1.00 L from 20.0 degrees C to 30.0 degrees C in a kettle. For the same amount of heat, how many kilo grams of 20.0 degree C air would you be able to warm to 30.0 degree C? What volume would this air occupy at 20.0 degrees C and a pressure of 1.00 atm? Make the simplifying assumption that air is 100% N2.



    2. Relevant equations

    m = 2CvQ/(3RΔTcv)

    ρ=m/v

    dQ = mcdT

    3. The attempt at a solution

    Q = (1.00kg)(4190 J/mol*K)(303K-293K) = 41900 J

    c(air) = 1.01 * 10^3 J/kg*K
    V= 1.00 L
    ρ(air) =1.2929 kg/m^-3
    m(air) = ρ(air)*V <=>
    m(air) = (1.2929 kg/m^-3)(1.00 L) = 1.2929 kg
    C(air) = (1.01*10^3 J/kg*K)(1.2929 kg) = 1305.829 J/mol*K

    R = 8.314 J/mol*K
    cv = 4190 J/mol*K

    the answer should be 5.65 kg
     
    Last edited: Jan 3, 2012
  2. jcsd
  3. Jan 3, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    Hint: you ignored V=const part.
     
  4. Jan 3, 2012 #3
    Is this formula right m = 2CvQ/(3RΔTcv) ?
     
  5. Jan 3, 2012 #4

    Borek

    User Avatar

    Staff: Mentor

    Argh, what was I thinking. You used wrong heat capacity value, but not because of the constant volume, you just have your units wrong. Sigh.

    Check your units. Is the result in kg?
     
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