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Homework Help: Thermodynamics and internal energy stuck

  1. Jan 7, 2006 #1
    alright, so I'm reviewing for a test I have on monday, and I was given a study guide which we must answer in order to get ready for the test. Now I just breeze through this stuff like its nothing, but I've found on discrepency. I even looked up the question online and i got 2 completely different answers, so, maybe someone here can guide me in the right direction, here it is.
    A thermodynamic system undergoes a process in which its internal energy decreases by 500 J. If at the same time, 220 J of work is done on the system, find the heat transferred to or from the system.
    its a simple dU = Q + W , but one place i found it ='s -720 [-500J - 220J] and the other I found it equaled -280 [-500J + 220J]. Any help is appreciated:smile:
  2. jcsd
  3. Jan 7, 2006 #2


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    Any work done on any isolated system increases its internal energy, my suggestion would therefore be that the -280J answer is correct because you taking out 500J but adding 220J back in.

    [EDIT] Thinking about it,it could be -720J. It is a rather ambigious question. Because if the system lost 500J but at the same time 220J of work were being done on the system, then the process would have to dissapate the 500J and a further 280J.

    It depends whether the 500J is the net internal energy decrease, but I can see both answer have a logical derivation. Sorry not much help there!
    Last edited: Jan 7, 2006
  4. Jan 7, 2006 #3
    You've probably thought right, but I think your notation is non-standard. It should be: U = Internal energy, W = mechanical work and Q = amount of heat added. dQ = dU + dW. All these changes are seen in relation to the system, so dU should be negative and dW positive, right?
    Last edited: Jan 7, 2006
  5. Jan 7, 2006 #4
    Hootenanny, doesn't "on the system" mean precicely that energy is added? Wouldn't it be "on the surrounding" if the system lost energy?
  6. Jan 7, 2006 #5
    oh yes, I see now. The one site that had the -720J answer notated it as

    3. DU = Q – W or Q = DU + W = –500J – 220 J = –720 J

    which would therefore yield that answer, I think... wow now I'm confusing myself, so it is -280J?
  7. Jan 7, 2006 #6


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    Yes but I was debating whether or not the 500J loss of energy included dissipating the energy from work done.
  8. Jan 7, 2006 #7
    I must say don't understand this reasoning. Concidering the wording in the question, why wouldn't the 500j be the net decrease in internal energy?
  9. Jan 7, 2006 #8
    alright, the formula is dU = Q - W. Work is done ON the system so W is going to be negative (right?). [W = -220J]. dU is -500J, because it initially lost that 500J so if you plug in ... -500J = Q -(-220). solving for Q would therefore yield Q = -720J. Am I correct???
  10. Jan 7, 2006 #9
    Yes, I think you are right... And I was wrong in saying dW would be positive if work is done on the system, it is positive when the system does work on the surrounding. Sorry.
  11. Jan 7, 2006 #10
    no problem thanks alot guys
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