Explore BrainMass

Explore BrainMass

    Maximum amount of work that can be extracted from two objects at different temperatures

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Consider a system consisting of two objects A and B, when A and B have initial temperature T_A and T_B, and heat capacities C_VA and C_VB independent of temperature. This system is put to work so that it can provide work W, until both reaches a final temperature T_F, where T_A>T_F>T_B.

    (a) Find the maximum mechanical work that can be extracted from this system.

    (b) Obtain the equilibrium temperature reached after all the work in (a) was extracted.

    © BrainMass Inc. brainmass.com December 24, 2021, 11:22 pm ad1c9bdddf
    https://brainmass.com/physics/heat-thermodynamics/maximum-amount-work-extracted-two-objects-different-temperatures-562994

    SOLUTION This solution is FREE courtesy of BrainMass!

    We can find the maximum amount of work that can be performed by considering the entropy of the system. If a system is not in thermal equilibrium, then allowing it to get into thermal equilibrium will lead to an entropy increase. If we extract work from that process, then the more work we extract from the system, the less the entropy will increase. This is because after we've extracted work, we are free to put back that work into the system in the form of heat. We would end up at the same final state where we would have arrived at, had we not extracted any work from the system. Obviously, the more work we put back into the system in the form of heat, the more the entropy will increase. Since we must end up at that same state which has some fixed entropy, this means that the state of the system after the work has been extracted will have a an entropy that will be lower if more work is extracted. The theoretical limit for the maximum amount of work that can be extracted is then obtained in the limit of zero entropy increase.

    So, what we need to do is find out what the entropy of the system is as a function of the temperatures of the two objects. The increase in the entropy of a system when an amount of heat of dQ is added to it in a quasi static way is given by dS = dQ/T. If the heat capacity is C, then this becomes:

    dS = C dT/T (1)

    In this problem, we want to assume a temperature independent heat capacity, but that's not possible when the temperature approaches zero. But we can assume that the heat capacity stays constant above some temperature T0 and we can assume that T0 < TA. If we integrate (1) to some temperature T > T0, we get:

    S(T) = Integral from T = 0 to T0 of dS + Integral from T = T0 to T of dS =

    S(T0) + C Integral from T = T0 to T of dT/T =

    S(T0) + C Log(T/T0) (2)

    Here we've used that S(0) = 0 (the third law of thermodynamics).

    Using (2) we can write the entropy of the system as:

    S(TA,TB) = SA(T0) + SB(T0) + C_VA Log(TA/T0) + C_VB Log(TB/T0) (3)

    If the final temperature of both objects is TF, then the entropy of the system will be:

    S(TF,TF) = SA(T0) + SB(T0) + C_VA Log(TF/T0) + C_VB Log(TF/T0) (4)

    We have zero entropy increase if S(TA,TB) = S(TF,TF):

    SA(T0) + SB(T0) + C_VA Log(TA/T0) + C_VB Log(TB/T0) = SA(T0) + SB(T0) + C_VA Log(TF/T0) + C_VB Log(TF/T0) ------>

    C_VA Log(TA) + C_VB Log(TB) = (C_VA + C_VB) Log(TF) --------->

    TF = TA^(CVA/C) TB^(C_VB/C)

    where C = C_VA + C_VB is the total heat capacity of the system.

    The amount of work that was extracted from the system follows from the change in the internal energy of the system. No heat was added to the system. While objects A and B will have had thermal interactions between each other during which energy was transferred in the form of heat between the objects, when we consider the system comprising of A and B together where A and B are then inside the system boundary, any energy that crossed that system boundary was in the form of work performed by the system. The internal energy function of the system can be written as:

    E(TA,TB) = EA(T0) + EB(T0) + C_VA (TA - T0) + C_VB (TB - T0)

    where like in case of the entropy, we have used that the heat capacity is constant for temperatures above some T0 < TA. If the system performs an amount of work W, then the internal energy decreases by W. This means that the work performed during the transition from the initial state to the final state where both temperatures TF, is given by:

    W = E(TA,TB) - E(TF, TF) = C_VA (TA - TF) + C_VB (TB - TF).

    Using the above expression for TF, we can write this as:

    W = C_VA TA + C_VB TB - C TA^(CVA/C) TB^(C_VB/C)

    = C[C_VA/C TA + C_VB/C TB - TA^(CVA/C) TB^(C_VB/C)]

    =C[arithimetic mean temperature weighted by heat capacity - geometric mean temperature weighted by heat capacity]

    Thank you for asking BrainMass.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 11:22 pm ad1c9bdddf>
    https://brainmass.com/physics/heat-thermodynamics/maximum-amount-work-extracted-two-objects-different-temperatures-562994

    ADVERTISEMENT