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QJEEM22Ph5a96: JEE Main 2022 Physics - Class 11 JEEM22Ph5a96 Solution: A block of ice of mass 120 g at temperature 0 ∘ C is put in 300 …

A block of ice of mass 120 g at temperature 0∘C is put in 300 g of water at 25∘C. The x g of ice melts as the temperature of the water reaches 0∘C. The value of x is _____________.

[Use specific heat capacity of water = 4200 Jkg−1K−1, Latent heat of ice = 3.5 × 105 Jkg−1]

Step-by-Step Verified Solution

Heat lost by water = Heat gained by ice

0.3×4200×25=x×3.5×105

x=0.3×4200×253.5×105

=90×100×105×103 gram = 90 gm

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