Verified Solution • 2025–26 Session moderate

QJEEM20Ph25dd: JEE Main 2020 Physics - Class 11 JEEM20Ph25dd Solution: The specific heat of water = 4200 J kg -1 K -1 and the latent…

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

Multiple Choice Options

  • A 63.8
  • B 61.7
  • C 69.3
  • D 64.6

Step-by-Step Verified Solution

Heat loss by water

Q=mwsΔθ

=(2001000).(4200)(25)=21000J

This heat will absorbed by the ice and let mass Δmi got melted.

So ΔmiL=21000

Δmi=210003.4×105×103gm=61.7grams

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