JEE Main20204 Sep 2020Morning ShiftPhysicsThermal Properties of MatterActual
The specific heat of water = 4200 J kg – 1 K – 1 and the latent heat of ice = 3.4 × 10 5 J k g – 1 . 100 grams of ice at 0 o C is placed in 200 g of water at 25 o C . The amount of ice that will melt as the temperature of water reaches 0 o C is close to (in grams)
Options
- A61.7
- B63 . 8
- C69 . 3
- D64 . 6
Correct answer
A. 61.7
Step-by-step solution
MSΔT = M l 200 1000 × 4200 × 25 = m × 340 × 10 3 m = 61 . 7