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How many kg of steam at 110 degree celsius is needed to have 10 kg of ice at 0 degree celsius converted to water at 30 degree celsius?




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Let  s1 = specific heat of steam.  in Joules/kg
     m1 = mass of steam in kg
     s2 = specific heat of water  in Joules/kg
     L1 = latent heat of steam  per kg
     L2 = latent heat of fusion.  per kg

heat given by steam when its temperature comes from 110 to 30 deg.C
   = heat absorbed by ice and converted water to rise to 30 deg C.

    s1 * (110 C - 100 C) * m1 + L1 * m1 + s2 * (100 C - 30 C) * m1
               = 10 kg * L2  +  10 kg * s2 (30 C - 0 C)

Substituting the other known values, you should the value of m1.

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