Question

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

Answer 1

Answer:

The value is $\eta = 54.4 \%$

Explanation:

From the question we are told that

    The work input is  $W = ( 245 + A ) \ J$

     The heat delivered is $Q = (142 + B) \ J$

      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

          $\eta = \frac{W}{Q_t}$

Here  $Q_t$ is the total out energy produce by the heat engine and this is mathematically represented as

           $Q_t= Q + W$

=>         $Q_t= 245 + A + 142 + B$

=>         $Q_t= 390 + A+B$

So

               $\eta = \frac{245 + A }{390 + A+ B}$

=>          $\eta = 0.544$

=>          $\eta = 0.544 *100$

=>          $\eta = 54.4 \%$