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Homework Help: Thermodynamics, Heat capacity.

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    The molar heat capacity at a constant volume of a solid at low temperatures T << Td, where Td is the Debye temperature is given by:

    [tex]C_v = 464(\frac{T}{T_d})^3 [/tex]

    Consider Td = 281 K for the NaCl.

    (a) Calculate the average molar heat capacity [tex]\bar{C_v}[/tex] of the NaCl between the temperatures of Ti = 10 K and Tf = 20 K.

    (b) Calculate the amount of heat necessary to raise the temperature of 1,000g of NaCl from 10 K to 20 K.

    2. Relevant equations

    [tex]C_v = 464(\frac{T}{T_d})^3 [/tex]

    [tex]C_v = (\frac{\delta Q}{dT})_v[/tex]

    3. The attempt at a solution

    (a) So I found the heat capacities Ci and Cf at Ti and Tf, respectively, and then in order to find the average heat capacity I summed them and divided by 2.

    [tex]C_i = 464(\frac{10}{281})^3 [/tex]

    [tex]C_i = 0,021 [/tex]

    [tex]C_f = 464(\frac{20}{281})^3 [/tex]

    [tex]C_f = 0,167 [/tex]

    [tex]\bar{C_v} = \frac{C_i + C_f}{2} [/tex]

    [tex]\bar{C_v} = 0,94 [/tex]

    Is this correct?


    [tex](\frac{\delta Q}{dT})_v = C_v [/tex]

    [tex]\delta Q = \bar{C_v} dT [/tex]

    [tex]\delta Q = (0,94)(10) = 9,4 [/tex]

    Is this also correct?

  2. jcsd
  3. Apr 26, 2010 #2
    I don't think the first part is correct since:

    f_{av} = \frac{1}{b-a} \int_{a}^{b} f(x) dx

  4. Apr 26, 2010 #3
    I see my mistake now, thank you very much kind sir.
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