Q7 of 41 Page 47

1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 105 J kg–1 and latent heat of vaporization of water = 2.26 × 106 J kg–1.

Given


Latent heat of fusion of ice = 3.36 × 105 J kg–1


latent heat of vaporization of water = 2.26 × 106 J kg–1


Formula used:


ΔQ=MCΔT


ΔQ=heat exchange


ΔT=tempeature change


M=mass


C=heat capacity


Heat lost 1 kg of steam= Heat gained by 1 kg of ice


3.36× 105+1× 1000× T=2.26× 106+1× 1000× (100-T)


This equation gives T>100 therefore we come to know steam is still present.


So making the temp of the mixture to be 100° C


3.36× 105+1× 4200× 100=M× 2.26× 106


M=0.3345kg


So mass of steam turned water=0.3345kg


So


Mass of water in mixture=1+0.3345=1.3345kg


Mass of steam left=1-0.3345=0.665kg


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