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How many kcal of heat are required to vaporize 25.0kg of water at 100°C?

  • Physics -

    1 J =0.2388 Cal
    Heat of vaporization for water
    λ = 2257•10^3•0.2388 =5.39•10^5 Cal/kg,
    Q=λ•m =5.39•10^5•25 =1.35•10^7 Cal =
    = 1.35•10^4 kCal

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