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Question

How much heat is required to convert 0.5kg of ice at −20∘C to water at 10∘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 = 0.5×2100×20 = 21000J (ice from -20°C to 0°C). Q2 = 0.5×3.35×105 = 167500J (melting). Q3 = 0.5×4186×10 = 20930J (water from 0°C to 10°C). Total: Q = 21000+167500+20930 = 208430J = 208.43kJ.

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