$A$ steam engine delivers $5.4 \times 10^{8} \;J$ of work per minute and receives $3.6 \times 10^{9} \;J$ of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
(N/A) Work done by the steam engine per minute,$W = 5.4 \times 10^{8} \;J$.
Heat supplied from the boiler,$H = 3.6 \times 10^{9} \;J$.
Efficiency of the engine $\eta = \frac{\text{Output energy}}{\text{Input energy}}$.
$\therefore \eta = \frac{W}{H} = \frac{5.4 \times 10^{8}}{3.6 \times 10^{9}} = 0.15$.
Hence,the percentage efficiency of the engine is $15 \%$.
Amount of heat wasted $= H - W = 3.6 \times 10^{9} - 0.54 \times 10^{9} = 3.06 \times 10^{9} \;J$.
Therefore,the amount of heat wasted per minute is $3.06 \times 10^{9} \;J$.
Heat supplied from the boiler,$H = 3.6 \times 10^{9} \;J$.
Efficiency of the engine $\eta = \frac{\text{Output energy}}{\text{Input energy}}$.
$\therefore \eta = \frac{W}{H} = \frac{5.4 \times 10^{8}}{3.6 \times 10^{9}} = 0.15$.
Hence,the percentage efficiency of the engine is $15 \%$.
Amount of heat wasted $= H - W = 3.6 \times 10^{9} - 0.54 \times 10^{9} = 3.06 \times 10^{9} \;J$.
Therefore,the amount of heat wasted per minute is $3.06 \times 10^{9} \;J$.
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View SolutionThe Carnot cycle (reversible) of a gas represented by a Pressure-Volume $(P-V)$ curve is shown in the diagram. Consider the following statements:
$I.$ Area $ABCD =$ Work done by the gas
$II.$ Area $ABCD =$ Net heat absorbed
$III.$ Change in the internal energy in the cycle $= 0$
Which of these are correct?
$I.$ Area $ABCD =$ Work done by the gas
$II.$ Area $ABCD =$ Net heat absorbed
$III.$ Change in the internal energy in the cycle $= 0$
Which of these are correct?
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