NTA Abhyas JEE Main2020PhysicsThermodynamicsPractice
The amount of heat energy required to freeze 4 . 5 g of water at 6 ° C to ice at 0 ° C is [ S = 4190 J kg - 1 K - 1 , L = 3.33 × 10 5 J kg - 1 ]
Options
- A1612   J
- B1512   J
- C1132   J
- D1499   J
Correct answer
A. 1612   J
Step-by-step solution
Total heat loss Q = m s ∆ T + m L ⇒     Q = ( 4.5 × 10 - 3 )   4190 6 - 0 + ( 4.5 × 10 - 3 ) ( 3.33 × 10 5 ) = 113 + 1499 = 1612   J