NTA Abhyas JEE Main2020PhysicsThermodynamicsPractice
In an energy recycling process, X g of steam at 100 ° C becomes water at 100 ° C which converts Y g of ice at 0 ° C into water at 100 ° C . The ratio of X Y will be (specific heat of water = 4200 J kg - 1 K , specific latent heat of fusion = 3.36 × 10 5 J kg - 1 , specific latent heat of vaporization = 22.68 × 10 5 J kg - 1 )
Options
- A1 3
- B2 3
- C3
- D2
Correct answer
A. 1 3
Step-by-step solution
Heat loss = Heat gain m 1 L v = m 2 .L f + m 2 .S . Δ T X × 10 - 3 × 22.68 × 10 5 = Y × 10 - 3 × 3.36 × 10 5 + Y × 10 - 3 × 4200 × 100 ∴ X Y = 7.56 22.68 = 1 3