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Practice question

MCQ

Question

How much heat is required to convert 0.4kg of ice at −25∘C to water at 50∘C? (Specific heat of ice = 2100J kg−1K−1, latent heat of fusion = 3.35×105J kg−1, specific heat of water = 4186Jkg−1K−1)

Options

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Explanation

Q1 = 0.4×2100×25 = 21000J (ice to 0°C). Q2 = 0.4×3.35×105 = 134000J (melting). Q3 = 0.4×4186×50 = 83720J (water to 50°C). Total: Q = 21000+134000+83720 = 238720J = 238.72kJ.

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