Practice question
Question
How much heat is required to convert 1kg of ice at −10∘C to water at 0∘C? (Specific heat of ice = 2100J kg−1K−1, Latent heat of fusion = 3.35×105J kg−1)
Explanation
Q1 = msΔT = 1×2100×10 = 21000J. Q2 = mLf = 1×3.35×105 = 335000J. Total heat: Q = Q1+Q2 = 21000+335000 = 356000J = 356kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 356 kJ. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.