Question

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 ...?

Answer 1

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.

Answer 2

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