Saved Bookmarks
| 1. |
A Carnot engine operates with source at 127^(@)C and sink at 27^(@)C. If the source supplies 40 kJ of heat energy the work done by the engine is |
|
Answer» Solution :The efficiency of a Carnot engine is `eta = 1 - (T_(2))/(T_(1))` Here, `T_(1) = 127^(@)C = 127 + 273 = 400 K` `T_(2) = 27^(@)C = 27 + 273 = 300 K` `:. eta = 1 - (300)/(400) = 1 - (3)/(4) = (1)/(4)` Also `eta = ("work done by engine")/("heat supplied by source") = (W)/(40 kJ)` `:. W = 40 eta = 40 xx (1)/(4) = 10 kJ` |
|