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Homework Help: Formulea in Heat and Thermodynamics

  1. Jan 9, 2006 #1
    I have forgotten most of the formulea in Heat and Thermodynamics. I would appreciate if some one can check the following for me please.

    1. A 2.2 Kg. block of ice melts to water at 0 degree celcius. By how much has the internal energy of the ice changed? Neglect the small change in volume.

    Q = M L where L is the latent heat of water = 3.33810^5 J/Kg
    Is this correct?

    2. Suppose that 2.3 mol of argon gas is heated from -45 degrees to 90 degrees celcius. Find the change in internal energy of the gas and the work it does if the heating occurs at a.) constant volume and b.) constant pressure

    At constant volume,

    work done, dW= p dV = 0

    From first law of TD,
    du = dq-w = dQ

    This is how I found dQ.
    dQ= M Cv (Tf - Ti) where M = number of moles * molecular weight of Ar.

    I found in the internet Cv(Ar)= 12.5 J/K at 15 degree celcius & 1 atm.
    Since Cv is a function of temperature, what value do I use for Cv in my problem?

    For constant pressure
    From pv=nRT, I have work done, W = PdV = nR dT = nR (T2-T1)
    To find the change in internal energy,
    du = Q - W = M Cp (T2-T1) - nR (T2-T1)
    I believe I can use Cp- Cv = R to find Cp.

    3. An average adult person generally emits energy at the rate of approximately 105 W while sedentary. At what rate is the persons entropy changing?
    No idea what formula to use. Any advise??

    4. Suppose you leave a 100 W light bulb on continuously for 'h' hours. If the electricity generated and delivered by the power company to the bulb at 30% efficiency, how much thermal energy does the power plant exhaust to the environment due to you oversight?

    Does this mean power plants loose 70% energy to the environment? :cry:
    This is how I worked out this problem

    Power * Time (in sec) * 70% = 180 MJ!! This seems so unreal. Am I right


    Last edited: Jan 9, 2006
  2. jcsd
  3. Jan 9, 2006 #2
    That looks fine, but remember you're solving for delta U, which equals Q, not Q itself.
    That seems right. Since you are dealing with argon, which is a monatomic gas that behves close to idealy, they may want you to assume c_v=3/2R. For a diatomic perfect gas this would be c_v=5/2R
    Alright, here you made a mistake. You have constant pressure and you want to integrate dW=PdV. You can just pull P outside the integral. You made an algebra mistake as well.

    To find delta U you do not have to find Q and W seperately.
    [tex]du=c_vdT+(\frac{\partial U}{\partial v})dv[/tex]
    But the second term equals 0 for a perfect gas so you could just integrate
    You'll get the same answer.
    There does not seem to be enough information to answer this question. I don't know what to tell you.
    That last answer isn't right. If you are getting 100W at 30% efficiency then they must be sending 100/.3=300W. 70% of this is lost to heat, or 210W. Now just multiply by time and convert to seconds.
    Last edited: Jan 9, 2006
  4. Jan 10, 2006 #3
    Thank You very much for your expert opinions.

    I can't understand where the mistake is. Did you mean that I have to plug in P = nRT/V in the integrant? I did not do this since Volumes are not given. But initial and finalt temps are given.

    dW = PdV
    W = int(P dV) = int (nR dT) = nR (T2-T1).

    Can you please point out where the algebra mistake is?

    I am trying to understand your answer to constant volume problem. As I said before I have forgotten all the formulea.

    Thank You.

  5. Jan 10, 2006 #4
    I didn't notice that you switched between dV and dT- sorry that was my mistake.
    These are just formulas. For an ideal monatomic gas:
    [tex]c_v=\frac{3}{2}R[/tex]; [tex]c_p=\frac{5}{2}R[/tex]
    For an ideal diatomic gas:
    [tex]c_v=\frac{5}{2}R[/tex]; [tex]c_p=\frac{7}{2}R[/tex]
  6. Jan 10, 2006 #5
    For the 3rd question, I figured the following.

    Rate of energy emitted by an average man is given. ie. dQ/dt.

    Can I do this?

    dQ = M Cv (T2 - T1)

    T2 = dQ/ MCv + T1 --------------(1)

    where M is the average man's mass and I would take
    T1 as his body temparature at normal conditions (~ 32 +273 K)

    Assuming constant volume, dQ = Cv dT = T dS

    ie. Cv dT = T dS

    Intergrating both side,

    delta(S) = Cv ln (T2/T1)

    Knowing T2 and T1, one can therefore find delta(S)

    What is Cv for a human?

    Does this look stupid way of solving the problem?
  7. Jan 10, 2006 #6
    Actually you can't do this. Q is not a state function. Imagine you had a container that was thermally isolated from the surroundings so that Q=0. A chemical reation could still go on that would raise the temperature of the system.
    You can't always write that dQ=TdS. Remember that the definition of entropy was
    Specifying the reversible path is very important. But I think you may be on to something with this. If the temperature of the surroundings is the same as the temperature of the bidy then the heating is reversible. Now, it appears that we are assuming the rate of heat transfer to be independent of the temperature, so the Q=Q_rev. This means that we can use the equaiton you gave. Just divide by dt (where t is time) to get the answer.
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