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482    CHAPTER 20 IRREVERSIBLE THERMODYNAMICS




                                                  l


                                                                  Q
                                (J ) T                          (J ) T+ T   J I
                                  Q


                                J' Q
             FIGURE 20.6
             Small element of wire subject to current flow.
             Let the required heat transfer, to a series of reservoirs, be J . Under these conditions, the heat flow due
                                                            0
                                                            Q
             to heat conduction is the same as previously.
                Then
                                  0
                                 J ¼     J Q  TþDT       J Q  T  þ   J I Dε              (20.61)
                                  Q
                                        |fflfflfflfflfflffl{zfflfflfflfflfflffl}  |fflffl{zfflffl}   |ffl{zffl}
                                      Heat flow rate; dQ=dt;  Heat flow rate;  Ohmic heating
                                          at TþDT       dQ=dt; at T
                At any given temperature, T, the heat of transport, Q*,is(Eqn (20.47))

                                                    J Q

                                              Q ¼        ¼ TS
                                                    J I
                                                        T
             and hence from Eqn (20.47)

                                                 J Q  ¼ J I TS                           (20.62)
                                                    T
                At a temperature, T þ DT

                                                             dS


                                   J Q     ¼ J I ðT þ DTÞ S þ   DT
                                      TþDT                   dT
                                                                                         (20.63)
                                TdS


                      ¼ J I TS þ    DT þ S DT   if multiples of small terms are neglected
                                 dT
                Now, for the element
                                                        dε
                                                Dε ¼     D‘                              (20.64)
                                                        d‘
             and
                                                        dT
                                                DT ¼      D‘                             (20.65)
                                                        d‘
                Equation (20.38) can be rearranged to give
                                               dε    J I    dT
                                                 ¼       S                               (20.66)
                                               d‘     l    d‘
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