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Homework Help: More Thermodynamics - please help =]

  1. Apr 9, 2005 #1
    A 50g block of copper having a temperature of 400K is placed in an insulating box with a 100g block of lead having a temperature of 200K.
    a) What is the equilibrium temperature of this two-block system?
    b) What is the change in the internal energy of the two-block system as it goes from the initial condition to the equilibrium condition?
    c) What is the change in the entropy of the two-block system?

    I set the heat loss of the copper block equal to the heat gain of the lead block to find the equilibrium temperature using the equation deltaQ=mcdeltaT and got 318K as the answer for part (a). Is that correct?

    Also, for part (b), would I find the change in internal energy of each block and add the two numbers together? Is there an equation to directly get changes in internal energy, or would I have to figure out the work first? (I know that deltaU = deltaQ - W)
  2. jcsd
  3. Apr 10, 2005 #2
    a) What is the heat capacity of the copper block and the lead block ? You cannot solve the problem without these two quantities. Heat gained by lead = heat lost by copper (in magnitude, ignoring s sign).

    b) First of all delta U = q + w for a closed system, not delta q - w. You can find the sum of the two quantities (remember to keep your signs right), but there is a much easier way to find the answer. Hint (how much heat is flowing in from the outside

    c) Whats your definition of entropy?
  4. Apr 10, 2005 #3
    a) Well, since you said that, is what I showed in my original post correct?

    b) There is no heat flowing in from the outside. So the change in internal energy would be zero?

    c) entropy = degree of disorder
    I have a general equation for change in entropy and then equations for isothermal, adiabatic, isochoric, and isobaric processes. Is this isochoric, since the volume inside the box does not change?
  5. Apr 10, 2005 #4
    a) I can't verify your numbers without the heat capacity of copper and lead.

    b) Yes . Since there is no heat flow and no work done from outside the change in internal energy is zero.

    c) Have you had calculus yet? dS = dQ/T. When you substitute dQ = Cp dT, and integrate this equation from some final to initial temperature. You wind up with delta S = Cp ln(Tf/Ti) But remember, that is for a single block so you will have to sum the two quantities.

    Since we are talking about solids and not gases, there really isn't a distinction made between an isobaric and isochoric processes, since solids don't expand with temperature very much, and pressure doesn't really have an effect on there thermodynamic behavor.
  6. Apr 10, 2005 #5
    a) oh sorry... for copper it is 386 J/kg*K and for lead it is 128 J/kg*K

    b) yay

    c) Okay, that makes sense. So I just use that equation for each block and add the two results together.

    And on the topic of the gases, for another problem there is a cycle (PV-diagram) where I have to calculate various things for each leg of the cycle. One part is an adiabatic process. Based on values I figured out for other parts of the cycle, I found that for an adiabatic process, the change in temperature is zero. So then that made the work and change in internal energy zero. That doesn't seem to make sense to me. So is that definitely wrong? If so, I need to go back and find where I went wrong with other things.
  7. Apr 10, 2005 #6
    Sorry to go a little off topic, this question brought up a question in me..

    I know dU = dQ + dW

    i.e, it's a process. How do you measure internal energy of of the 50g of copper block at 400K in this case? And does this equation only apply to gases?

  8. Apr 10, 2005 #7

    I can't really know what went wrong without the question but anyway hopefully this can be of some help to you.

    In gases, the first law states that,

    dU = dQ + dW.

    In an adiabatic process, is one which takes place such that no heat flows in or out of the system, so dQ = 0. First Law becomes

    dU = -dW. (-ve because work is done by gas)

    This implies that work is done at the expense of internal energy of the system.

    ===> dW > 0 (adiabatic expansion
    dU < 0 (internal energy decreases)
  9. Apr 10, 2005 #8
    Okay, so then most definitely the work done and change in internal energy CANNOT be zero?

    Well I'll have to go through the other parts of the problem again. Hopefully I'll figure out where I went wrong.
  10. Apr 10, 2005 #9
    P.S. THANKS for all the help. I'm going to bed now but if you have anymore input I'll definitely check back in the morning.

    Thanks again!!
  11. Apr 10, 2005 #10
    No, this expression applies to all systems. It is a fundamental law of thermodynamics (the First Law). You cannot intrinsically measure the internal energy of a system. Only changes are meaningful, and those changes are the heat added or lost plus and the work done on or by the system. Heat and work are actually just two forms of kinetic energy, so what you are really measuring is the net change in kinetic energy of the entire system. Heat is a more "random" type of energy, and so it reasonably follows that it is a less usable form of energy, and from there you can see the justification for the Second Law of Thermodynamics.
    Last edited: Apr 10, 2005
  12. Apr 10, 2005 #11
    No, internal energy is unchanged in the SEALED, ISOLATED container. Internal energy changes for the copper block, the lead block, and the air in between them(assumed to be unimportant), but the system as a whole, has no net change in internal energy. Because work is not done, and heat does not flow out of, the container. The container cannot expand, so it cannot do expansion work. The pressure may increase or decrease, but no work can be done since dw = - p dV and dV = 0. The lead block will expand slightly, and the copper block will shrink slightly, but the question doesn't ask you to calculate the work done for those components.
  13. Apr 10, 2005 #12
    Thanks for the clarification so-crates. That piece of information was based on flythisforme's question on gases. But thanks once again for the help on the first law.
  14. Apr 10, 2005 #13

    Based on adiabatic process itself, the change in Internal Energy of the system (dU) is dependent on work done BY or ON the gas (dW). So, work done by or on the gas cannot be 0, since no heat was flows in or out of the system, unless dU is 0.
    Last edited: Apr 10, 2005
  15. Apr 10, 2005 #14
    So it is possible for dU to be zero in an adiabatic process? And then the work would be 0? Because I am getting zero temperature change, and this causes the deltaQ, deltaW, and deltaU to be zero.

    [EDIT] Maybe I should clarify a bit. I guess what I am asking is that can a process be adiabatic AND isothermal?
    Last edited: Apr 10, 2005
  16. Apr 11, 2005 #15
    If dU is 0, it means that there is no change in internal energy of the system, i.e, dW is also 0, which implies not doing anything to the system at all.

    Adiabatic processes will definately result in a change in temperature. Since dU = dW, when work is done by the gas, dU is -ve -> final temperature is lower than the initial, since total internal energy decreases in the system.

    When work is done on the gas, dU is +ve -> final temperature is higher than initial, since total internal engery increases in the system.

    To justify this, look to the topic on ideal gases,

    1/2 N m <c^2> = 3/2 N K T

    where N is no. of molecules, m is mass of one molecule, <c^2> the mean square speed of the molecules, K is boltszmann constant and T is temperature is Kelvins.

    So, since 1/2 N m and 3/2 N K are constants, K.E of gas is directly proportional to Temp.

    Hope this helps.
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