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A Carnot engine operating between temperatures T_(1) and T_(2) has efficiency (1)/(6). When T_(2) is lowered by 62 K its efficiency increase to (1)/(3). Then T_(1) and T_(2) are, resectively : |
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Answer» 372 K and 330 K `rArr (T_(2))/(T_(1))=(5)/(6)"….(i)"` `eta_(2)=1-(T_(2)-62)/(T_(1)=(1)/(3)"….(ii)"` Solving we GET, `T_(1)=372 K and T_(2)=(5)/(6)xx372=310 K` Correct choice : (d). |
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