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The specific heat of water =4200 J kg^–1 K^–1 and the latent heat of ice =3.4× 10^5 J kg^–1. 100 grams of ice at 0 ^ o C is placed in 200 g of water…

JEE Main 2020 — Physics Thermodynamics

2020mcqmedium

The specific heat of water =4200 J=4200\text{ J}=4200 J kg–1 K–1{\text{kg}}^{–1}{\text{ K}}^{–1}kg–1 K–1 and the latent heat of ice =3.4×105 J=3.4\times {10}^{5}\text{ J}=3.4×105 J kg–1{\text{g}}^{–1}g–1. 100100100 grams of ice at 0oC0{ }^{\text{o}}\text{C}0oC is placed in 200 g200\text{ g}200 g of water at 25oC25{ }^{\text{o}}\text{C}25oC. The amount of ice that will melt as the temperature of water reaches 0oC0{ }^{\text{o}}\text{C}0oC is close to (in grams)

Official previous-year question

Held on 4 Sept 2020 · Verified 6 Jul 2026.

Options

  1. A

    61.761.761.7

  2. B

    63.863.863.8

  3. C

    69.369.369.3

  4. D

    64.664.664.6

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