The specific heat of water = 4200  J kg – 1  K – 1 and the latent heat of ice = 3.4…

The specific heat of water =4200 J kg–1 K–1 and the latent heat of ice =3.4×105 J kg–1. 100 grams of ice at 0 oC is placed in 200 g of water at 25 oC. The amount of ice that will melt as the temperature of water reaches 0 oC is close to (in grams)
  1. 61.7
  2. 63.8
  3. 69.3
  4. 64.6

Solution

MSΔT=Ml

2001000×4200×25=m×340×103

m=61.7

Asked in: JEE Main 2020 (04 Sep Shift 1)

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