2 kg of ice at - 20°C is mixed with 5 kg of water at 20°C . The water content of the final mixture is (Latent heat of ice = 80 kcal kg - 1 , the specific heat of water = 1 kcal kg - 1 ° C - 1 and specific heat of ice = 0.5 kcal kg – 1 ° C – 1 )

Question

2 kg of ice at - 20°C is mixed with 5 kg of water at 20°C . The water content of the final mixture is (Latent heat of ice = 80 kcal kg - 1 , the specific heat of water = 1 kcal kg - 1 ° C - 1 and specific heat of ice = 0.5 kcal kg – 1 ° C – 1 ), 7 kg, 6 kg, 4 kg, 3 kg

MARKS App · Accepted Answer

Here, m , s and θ 1 are the mass, specific heat and temperature difference for water respectively. Similarly, M , S and θ 2 are the corresponding quantities for ice. Here, L is the latent heat of fusion. We have, the heat lost by water = msθ , heat gained by ice = MSθ + ML . From the principle of calorimetry, we have, the heat lost by water=heat gained by ice. Hence, msθ 1 = MSθ 2 + ML . 5 × 1 × 20 - 0 = 2 × 0 . 5 × 20 + x × 80 x = 1 kg 1 kg of ice melts to form water. Hence, final amount of water 5 kg + 1 kg = 6 kg .

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