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Heat supplied to a Carnot engine is `453.6 kcal`. How much useful work can be done by the engine that works between `10^(@)C` and `100^(@)C`? |
Answer» `T_(2)=100+273=373K,T_(1)=10+273=283K`, `q_(2)=453.6xx4.184=1897.86kJ` We know that, `w=q_(2)*(T_(2)-T_(1))/(T_(2))` `=1897.86xx((373-283))/(373)` `=(1897.86xx90)/(373)=457.92kJ` |
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