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A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find the efficiency of the heat engine. Give your answer in percent (%) and with 3 significant figures.
A=14, B=72

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okpalawalter8

Answer:

The value is [tex]\eta = 54.4 \%[/tex]

Explanation:

From the question we are told that

    The work input is  [tex]W = ( 245 + A ) \ J[/tex]

     The heat delivered is [tex]Q = (142 + B) \ J[/tex]

      The value of A is  A =  14

        The value of B  is  B  = 72

Generally the efficiency of the heat engine is mathematically represented as

          [tex]\eta = \frac{W}{Q_t}[/tex]

Here  [tex]Q_t[/tex] is the total out energy produce by the heat engine and this is mathematically represented as

           [tex]Q_t= Q + W[/tex]

=>         [tex]Q_t= 245 + A + 142 + B[/tex]

=>         [tex]Q_t= 390 + A+B[/tex]

So

               [tex]\eta = \frac{245 + A }{390 + A+ B}[/tex]

=>          [tex]\eta = 0.544[/tex]

=>          [tex]\eta = 0.544 *100[/tex]

=>          [tex]\eta = 54.4 \%[/tex]

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