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Efficiency of engine is eta_(1)" at T_(1)= 200^(@)C and eta_(2)" at "T_(1)=0^(@)C and T_(2)=-200 K. Find the ratio of (eta_(1))/(eta_(2)) : |
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Answer» `1 cdot 00` `eta_(2)=1-((-200+273))/(273)=(200)/(273)` `therefore (eta_(1))/(eta_(2))=(273)/(473)=0 cdot 577` So, correct choice is (b). |
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