Page 125 - Essentials of physical chemistry
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The Second and Third Laws of Thermodynamics                                  87

            valve parts of an iron engine as just below the melting point of iron at about 9258K, we can estimate
            a maximum efficiency of about 60%:

                                          2300   925
                                                    ¼ 0:598 ffi 60%:
                                     E eff ¼
                                             2300
            In practice this is rarely attainable, but even this idealized efficiency shows us that practical heat
            engines waste a lot of heat energy on the T l part of the cycle. Due to internal friction of the moving
            parts and other design considerations, an actual efficiency of 30% is good. These examples illustrate
            the power of using the Carnot efficiency formula for any heat engine. All we need to know is the
            operating range of temperatures in absolute K degrees, but we should realize this is usually an upper
            limit to the efficiency of a real engine.


            ENTROPY AND SPONTANEITY
            From life experience, we know that heat tends to flow from hot to cold unless some sort of
            refrigeration device is used. Although the Carnot cycle established the existence of entropy, let us
            give some consideration to the sort of simple analysis Clausius would give for a heat engine.
            Consider a ‘‘heat source’’ at T hot that supplies heat energy to an ‘‘engine’’ so that the heat source
            provides jqj to the engine, which then discards jq j heat to the ‘‘environment’’ at T cold . We choose to
                                                   0
            use the absolute values jqj and jq j as positive numbers with the algebraic signs to indicate which
                                       0
            way the heat flows. The work the engine does is thus the difference in the energy (jqj jq j).
                                                                                        0
            A simple analysis gives

                                                                 jqj
                                                                    ,
                                 Hot source loses heat jqj, so DS 1 ¼
                                                                 T hot
                                                                    0
                                Cold source gains heat jq j,  so DS 2 ¼  þjq j  ,
                                                     0
                                                                 T cold
                                                                 0
                                                                   :
                                                          jqj  jq j
                                   Total DS ¼ DS 1 þ DS 2 ¼  þ
                                                         T hot  T cold
            Next, we use the idea that ‘‘spontaneous heat flow’’ occurs as long as DS > 0. We have to think
            about this, but we know heat flows from hot to cold and the absolute temperature in the denominator
            is always positive so the sign of DS depends on the heat flow. When the heat flow stops, the DS
                                                                          0
            value will be zero. Thus, the heat flow will just stop (DS ¼ 0) when   jqj  þ  jq j  ¼ 0, which can be
                                                                  T hot  T cold

                                T cold   jq j  T cold
                                          0
                                     and         . If we have to discard jq j heat energy into the cold
                        0
                                                                     0
            rearranged to jq j¼jqj          ¼
                                T hot    jqj  T hot
            sink (exhaust), the maximum energy that can be extracted as work is jqj jq j. Now define the
                                                                            0
            efficiency of the engine as how much work can be obtained for the input heat absorbed:
                              work produced  jqj jq j     jq j     T cold  T hot   T cold
                                                   0
                                                            0
                                                                                   :
                               heat absorbed    jqj        jqj     T hot    T hot
                   Efficiency ¼            ¼         ¼ 1     ¼ 1       ¼
            Thus, we obtain the same formula for the efficiency of a heat engine as from the Carnot cycle. Along
            the way, we realized that when heat flows from hot to cold, the environment will gain the exhaust
            heat jq j so that the environment gains entropy. A profound result of this sort of analysis is that
                 0
            entropy tends to increase in the environment unless there is some other condition and the overall
            entropy in the universe tends to increase. Especially for biology majors and generally for all of us,
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