InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
In a refrigerator, one removes heat from a lower temperature and deposites to the surrounding at a higher temperature. In this process, machanical work has to be done, which is provided by an elecrtic motor. If the motor is of `1KV` power, and heat is transferred from `-3^(@)C "to" 27^(@)C`, find the heat taken out of the refrigerator per second assuming its efficiency is `50%` of a perfect engine. |
|
Answer» Here, power of motor, `W= 1kW` `T_(1)= 27^(@)C= (27+273)K= 300 K, T_(2)= -3^(@)C= (-3+273)K= 270K` `eta = 1-(T_(2))/(T_(1))= 1-(270)/(300)= 1/(10)`. Efficiency of refrigerator `= 0.5 eta= 1/(20)` `COP= beta=(Q_(2))/W= (1-eta)/(eta)= (1-1/20)/(1//20)=19` `Q_(2)= 19W = 19W :.` Heat taken out of refrigerator/sec `= Q_(2)= 19kW` |
|