By Michael Sprackling (auth.)
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Example text
With the given restrictions the internal energy U of the system is unchanged by the process; that is, flU= 0 flU is given by flU= flU(lead) + flU(copper) + flU(water) where flU(lead) is the change in the internal energy of the lead, and so on. Because it is assumed that there are no work interactions, for each part of the system the change in the internal energy is equal to the heat absorbed. For example, flU(lead) = Q(lead) where Q(lead) is the heat absorbed by the lead during the process, and is given by Q(lead) = m(lead) X cP(lead) X (Tf - T;)(lead) where m is the mass, cP is the specific heat capacity at constant pressure, T; is the initial temperature and Tf is the final temperature.
However, at the start and end of the process the gas is in an equilibrium state in which the coordinates are well-defined quantities and relationships may be looked for between the coordinate values in these states. Let the system undergo an infinitesimal free expansion. The change in U may be written as a function of any two other coordinates and it is convenient to choose temperature and volume. Then dU = (amav)T dV + (amaT)v dT Now, in any infinitesimal free expansion dU is zero and Joule's experiment suggests that, for an ideal gas, dT is also zero.
Neglect the work done on the bodies by the surrounding atmosphere. Solution Let the vessel with adiabatic walls constitute the boundary of the system, which therefore comprises the two bodies that will be brought into thermal contact. Since the boundary is adiabatic there is no thermal interaction between the system and its surroundings. Further, the work interaction may be neglected. U2 that of body 2. The first law of thermodynamics may also be applied to each body separately. Then and where Q1 is the heat absorbed by body 1 and W1 is the work done on body 1 during the process.
