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What mass of steam at 100 C must be mixed with 150 gram of ice at its melting point, in a thermally insulated container, to produce liquid water at 50 C?


Heat of fusion of water at its normal freezing or melting temperature = 333 kJ/kg

Heat of vaporization (or steam) = 2256 kJ/kg

Specific heat of water = 4180 J/kg K ...?

Verified Answer :

vanessa61
  1. Since the system is thermally insulated, then the 

    Heat given up by steam = Heat absorbed by ice 

    Qs = Qi 

    Qs = Ms(Hv) + Ms(Cp)(100 - 65) 

    Qs = Ms(Hv) + Ms(Cp)(35) 

    where 

    Ms = mass of steam 
    Hv = heat of vaporization of steam 
    Cp = specific heat of water 

    Qi = 0.350(Hf) + 0.350(Cp)(65 - 0) 

    Qi = 0.350(Hf) + 22.75(Cp) 

    where 

    Hf = heat of fusion of ice 
    Cp = specific heat of water (as previously defined) 

    and since Qs = Qi, then 

    Ms(Hv) + Ms(Cp)(35) = 0.350(Hf) + 22.75(Cp) 

    I will stop my actual solution at this point. From hereon in, I trust that you can proceed with the rest of the solution. 

    Simply, determine the following values -- Hv, Cp and Hf -- and substitute in the above equation. After substituting the appropriate values, then you can already solve for Ms -- the mass of steam. 






    2.      m (2256 + 51 x 4.186) = 485 (333 + 49 x 4.186) 
    m = 105.684 g

scme1702

The ice will require two forms of heat: latent to melt and sensible to be heated to 50 °C.
Q(ice) = ml + mCpΔT
= 150 x 333 + 150 x 4.18 x 50
= 85950 Joules
The mass of steam must release this much energy in two forms: latent to fuse into water and then sensible to cool to 50 °C.
85950 = m(2256) + 4.18 x 50 x m
m = 34.9 grams of steam.

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