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Adiabatic electrical Thermodynamics question - Please Help

  1. Nov 6, 2005 #1
    Adiabatic electrical Thermodynamics question - Please Help!!

    The question is a thermodynamics question where the system is a resistor in a circuit; the question is as follows:
    A thermally insulated (Adiabatic) resistor of 30 Ohm's has a current of 2 Amperes passed through it for 1 second. It is initially at 20 degrees celcius (= 293 degrees K). The mass, m of the resistor is 6 grams. If Cp (specific heat capacity at constant pressure) = 850 J/kg-K calculate :
    a). The temperature increase, and
    b). The change in entropy (Sf - Si) of the resistor and the universe.
    (Hint: The actual process is dissipative. Imagine a reversible process taking it between the 2 equilibrium states.)

    Please help :eek:
     
  2. jcsd
  3. Nov 6, 2005 #2

    Pengwuino

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    Gold Member

    What have you done so far?
     
  4. Nov 6, 2005 #3
    well, adiabatic process means dq = 0. thus dU = -dW. Now V = I*R, and thus V = 60 volts. i forgot how to convert this to joules, i divided by the time of 1 second, but im not sure on that. up to there on this part, then for part b, which you need the temp increase from a to solve it, i worked out and left it in variable form since i didnt have the value for T2 (final temp). Here is what i did for this part:
    T dS = Cp*dT => dS = Cp * (dt/T) => delta-S = Cp * [ln(T2/T1)].
    Now, at this point, i am not sure why i did it, but i multiplied this by the mass, and it yields:
    850 J/kg-K * 0.006 kg * ln(T2/T1) = 5.1 * ln(T2/T1)
    with T1 from part a). you can find T2. but i think this is pretty wrong, i might be somewhat on the right path, but there is definitely an error along the way. I am going to look up the formula that relates the voltage and electrical info to the work. Any tips or pointers?
     
  5. Nov 6, 2005 #4
    so do u have any ideas where to go with this?
     
  6. Nov 6, 2005 #5

    Pengwuino

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    Well voltage * current in a dc application gives you the power and since it is only for one second, it gives you the joules as the same value.

    If you have the mass and the specific heat and the energy, you can use...

    [tex]\Delta E = mc\Delta T[/tex]

    To determine the temperature change.

    I forget how to do entropy change however, I'm sure there is an equation that tells you how to do that as well.
     
  7. Nov 8, 2005 #6
    thanks

    i got it, and so u know, the last part for entropy would be:

    once u have Change in Temp., which is 25.53 K, we have the expression

    T*dS = Cp*dT​

    and this becomes Delta*S = Cp*ln(T2/T1)
    where T1 = 293.15 and T2 = 316.68

    but, Cp that is given is in terms of specific quatities, i.e. Cp = 850 J/kg*K, and therefore you must multiply by the mass of the resistor, 0.006 kg, to obtain the non-specific Entropy S, which turns out to be

    0.394 J/K​
    thanks again :biggrin:
     
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