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A Carnot engine takes in heat from a reservoir of heat at `427^(@)C`. How many calories of heat must it take from the reservoir in order to procuce useful mechanical work at the rate of `357 watt`? |
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Answer» Here, `T_(1)= 427^(@)C = 427+273` `=700K` `T_(2)= 77^(@)C= (77+273)K= 350K`, `Q_(1)=?, W=357 "watt"`, As `(Q_(2))/(Q_(1))= (T_(2))/(T_(1))` `:. (Q_(1)-W)/(Q_(1))= (350)/(700)= 1/2` `Q_(1)-W= (Q_(1))/2 or (Q_(1))/2=W`, `Q_(1)= 2W=2xx357= 714 "watt"` `=(714)/(4.2)cal//s= 170 cal//s` |
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