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I Is there an Entropy difference between a cold and hot body

  1. May 5, 2018 #1
    Hello,

    If two bodies, who say start with ##T_{cold}=T_c## and ##T_{hot}=T_h## and then they are brought in contact with one another and then after some time they both have the same temperature. What would be the entropy of the entire system?

    Also another quick question, I've looked at some example problems of entropy and I would like to know why sometimes the entropy can be negative? For example, 1000J of heat is used into 470J of work when ##T_1 = 540K## and ##T_2=300K##.

    ##\triangle S = \frac{-1000J}{540K} = -1.85J/K## Why do we have a minus sign?
     
  2. jcsd
  3. May 5, 2018 #2
    Are you asking about entropy changes, or are you asking about absolute entropy values?
     
  4. May 6, 2018 #3
    The first question is what would be the entire entropy of the system after such a process occurs.
     
  5. May 6, 2018 #4
    cold body will have entropy increased. hot body will have entropy decreased. In the process of thermal conduction, there is new entropy added. I am not sure about the name of such new entropy, maybe it is called entropy of mixing (not sure).

    If you try to calculate the entropy change for both bodies after thermal conduction, you will find out that every time there is thermal conduction, entropy of the system will increase. If you try to reduce entropy of an object, you are bound to increase entropy somewhere in the system. For example, you try to divide some different objects into groups, the objects will have entropy reduction due to you, but you as an inanimate object will have a lot of entropy increase. For example, the reaction inside your body. To make a small reduction of a "local entropy", you need a bigger increase in "global entropy".

    Therefore, for a system, entropy will increase after thermal conduction.
     
  6. May 6, 2018 #5
    For the first problem, let's concentrate on determining the change in entropy between the initial state (where both bodies are at different temperatures) and the final state (in which both bodies have equilibrated to the same final temperature). What have you learned so far regarding how to determine the change in entropy of a system between an initial state and a final state?
     
  7. May 6, 2018 #6
    I learned that entropy is a state quantity and for irreversible process the total entropy is the initial state minus the final state.
     
  8. May 6, 2018 #7
    I think by writing that bit I answered my second question regarding the negative sign. But the first question I'm still keen on delving into.
     
  9. May 6, 2018 #8
    What is the equation they gave you in your course for calculating the change in entropy of a system (hint: it involves a quantity with the symbol ##q{rev}##)?
     
  10. May 6, 2018 #9

    Mister T

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    All we can determine is that if the system you describe is closed, its entropy increases

    This minus sign tells you that ##\Delta S## is negative, meaning the entropy ##S## has decreased.

    What this calculation tells me is that 1000 J of heat was transferred from (rather than to) a system while that system maintained a constant temperature of 540 K.
     
  11. May 6, 2018 #10
    This is definitely not connect. Not only can we quantify the change in entropy for this overall system, but we can also quantify the change in entropy for each of the bodies individually.
     
  12. May 7, 2018 #11
    The equation is ##d S = \frac{dQ_{rev}}{T}##

    I would also like to assume that the system doesn't lose/gain any heat from the outside. I.e. the system is closed. So is the change of Q = 0 and the temperature of the system is constant.

    Usually ##\triangle S = Q_1/T_1 + Q_2/T_2## but in this case I don't see how we can use this formula since we don't know the quantity of heat and we have two objects so we would have to have ##\triangle S = Q_1/T_{1,hot} + Q_2/T_{2,warm} +Q_1/T_{1,cold} + Q_2/T_{2,warm} ##
     
    Last edited: May 7, 2018
  13. May 7, 2018 #12
    I can help you through all these difficulties to get to the final answer. The first step in the process is to use the first law of thermodynamics to determine the temperature of the two bodies in the final state ##T_F##. Without knowing the final state, we can't get the entropy change, since entropy is a function of state.

    So if the mass, heat capacity (per unit mass), and temperature of the initially hot body are ##m_H##, ##C_H##, and ##T_H## (respectively), and the mass, heat capacity (per unit mass), and temperature of the initially cold body are ##m_C##, ##C_C##, and ##T_C## (respectively), then algebraically, in terms of these parameters, what is the final temperature of the two bodies (after they have equilibrated)?
     
  14. May 7, 2018 #13
    The final temperature would be ##T_W ##(warm). The hot body would decrease temperature i.e. give heat to the cold body and the cold body would increase temperature absorbing heat from the hot body.
     
  15. May 7, 2018 #14
    Do you now know how to determine this algebraically in terms of the variables I specified (using the 1st law of thermodynamics)?
     
  16. May 7, 2018 #15
    I figured that the system must be isochor since the volume stays the same and there is not work being done to the system. So ##\triangle Q = \triangle U = c_vM\triangle T = \frac{f}{2}Nk\triangle T## but ##C_v## is the molar specific heat, right? ##c_v = \frac{fk}{2m}## and ##C_v = \frac{f+2}{R}##
     
  17. May 7, 2018 #16
    This is only if the "bodies" are ideal gases. Assuming that the bodies are solids, please answer my question regarding the final temperature.
     
  18. May 7, 2018 #17
    Ah right. No unfortunately I do not.
     
  19. May 7, 2018 #18
    Does this equation make any sense to you: $$T_F=\frac{(m_HC_HT_H+m_CC_CT_C)}{(m_HC_H+m_CC_C)}$$
    This is what you get if you set the change in internal energy of the combination of the two bodies equal to zero.
     
  20. May 7, 2018 #19
    Where does this equation exactly come from? Why would the internal energy be zero when there is internal energy in the system? The difference in the internal energy is zero but nonetheless there is internal energy.
     
  21. May 7, 2018 #20
    Maybe this will make more sense. The change in internal energy of the hot body will be
    $$\Delta U_H=m_HC_H(T_F-T_H)$$
    The change in internal energy of the cold body will be
    $$\Delta U_C=m_CC_C(T_F-T_C)$$
    The change in internal energy of the combined system of cold body and hot body will be zero.
    $$\Delta U=\Delta U_H+\Delta U_C=0$$
     
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