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 J kg - 1 K - 1 , Latent heat of ice = 3 . 5 × 10 5 J kg - 1 ]

Question

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 J kg - 1 K - 1 , Latent heat of ice = 3 . 5 × 10 5 J kg - 1 ],

MARKS App · Accepted Answer

According to the principle of calorimetry, heat lost by hot body is equal to the heat gained by cold body. The heat released by water ∆ Q = m s ∆ T = 0 . 3 × 4200 × 25 = 31500 J let m kg ice melts then, ∆ Q = m L = m × 3 . 5 × 10 5 = 31500 J ⇒ m = 31500 × 10 - 5 3 . 5 = 9000 × 10 - 5 m = 0 . 09 kg = 90 g ⇒ x = 90

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