NTA Abhyas JEE Main2020PhysicsThermodynamicsPractice
A refrigerator absorbs 2000   c a l of heat from ice trays. If the coefficient of performance is 4 , then work done by the motor is 1   cal = 4 . 2   J
Options
- A2100 J
- B4200 J
- C8400 J
- D500 J
Correct answer
A. 2100 J
Step-by-step solution
Q 2 = 2000   cal Coefficient of performance, COP = Q 2 W W = Q 2 COP = 500   cal W = 500 × 4 .2   J = 2100   J