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A Carnot's engine is working between 7^(@)C and 287^(@)C. It is desired to increase its efficiency to 70%. By how much should the temperature of source be increase ? |
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Answer» `273^(@)C` `ETA =(70)/(100) cdot" then "(7)/(10)=(T_(1)-280)/(T_(1))" or "T_(1)=933.3 K.` INCREASE in temperature `=933.3-560=373.3^(@)C`. `therefore` Correct choice is (c ). |
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