Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 15 - Thermodynamics - Problems - Page 417: 47

Answer

5% more work will be done by engine after tuning.

Work Step by Step

Suppose before tuning the engine efficiency was $e_1$ Heat input is given as $Q_1=1300J$ suppose the work done is $W_1$ from definition of efficiency of engine $e_1=\frac{W_1}{Q_1}$ So $W_1=e_{1}Q_{1}$ $W_1=1300e_{1}J$.....................(1) suppose after tuning the engine the efficiency is $e_2$ given that new efficiency is 5% more than old efficiency $e_2=e_1 +\frac{5e_1}{100}$ suppose in this case work done by the engine is $W_2$ Heat input is same as $Q_1=1300J$ from definition of efficiency of engine $e_2=\frac{W_2}{Q_1}$ so work done by the engine after tuning is $W_2=e_{2}Q_{1}$ so $W_2=(e_1 +\frac{5e_1}{100})1300J$................equation(2) difference in work done after tuning increase in work done $\Delta W=W_2-W_1$ $\Delta W=(e_1 +\frac{5e_1}{100})1300J-1300e_{1}J$ amount of work increased $\Delta W=1300J\times\frac{5e_1}{100}$ so % increase in work $\Delta W$%=$\frac{\Delta W}{W_1}\times100$ $\Delta W$%=$\frac{1300J\times\frac{5e_1}{100}}{1300e_{1}J}\times100$ % $\Delta W$%=$5$% so 5% more work will be done by engine after tuning.
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