problem20_62

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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20.62: a) As in Example 20.10, the entropy change of the first object is ) ln( 1 1 1 T T c m and that of the second is ) ln( 2 2 2 T T c m , and so the net entropy change is as given. Neglecting heat transfer to the surroundings, , 0 ) ( ) ( , 0 2 2 2 1 1 1 2 1 = - + - = + T T c m T T c m Q Q which is the given expression. b) Solving the energy-conservation relation for T and substituting into the expression for S gives . 1 n 1 ln 2 1 2 2 2 1 1 2 2 1 1 1 - - + = T T T T c m c m c m T T c m S Differentiating with respect to T and setting the derivative equal to 0 gives - - - + = 2 1 2 2 2 1 1 2 2 2 1 1 2 2 1 1 ) ( 1 ) 1 )( )( ( 0 T T T T c m c m T c m c m c m T c m . This may be solved for , 2 2 1 1 2 2 2 1 1 1 c m c m T c m T c m T + + = which is the same as T when substituted into the expression representing conservation of energy. Those familiar with Lagrange multipliers can use that technique to obtain the relations
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Unformatted text preview: T Q S T T Q S T ′ ∂ ∂ = ∆ ′ ∂ ∂ ∂ ∂ = ∆ ∂ ∂ λ , λ and so conclude that T T ′ = immediately; this is equivalent to treating the differentiation as a related rate problem, as 2 2 1 1 = ′ ′ + = ∆ ′ dT T d T c m T c m S T d d and using T T c m c m dT T d ′ =-= ′ gives 2 2 1 1 with a great savings of algebra. c) The final state of the system will be that for which no further entropy change is possible. If , T T ′ < it is possible for the temperatures to approach each other while increasing the total entropy, but when , T T ′ = no further spontaneous heat exchange is possible....
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