What is the result of mixing 1kg ice at 0°c with 100kg steam at 100°c? Solve

1 Answer
May 25, 2018

Compare Latent heat of fusion of ice=3.36 × 10^5\ Jcdotkg^-1
and
Latent heat of vaporization of water = 2.26 × 10^6\ Jcdotkg^-1

We see that steam will condense to become water at 100^@C and Ice will melt to become water at 0^@C

Heat required to melt 1\ kg of ice to become water at 0^@C
=3.36 × 10^5\ J
This much heat can be given by amount of steam=(3.36xx10^5)/(2.26xx10^6)
=0.151\ kg

Heat required to raise temperature of 1\ kg of water at 0^@C to 100^@C =msDeltaT=1xx4186xx100=4.186xx10^5\ J
(Specific heat of water s=4186\ J cdot kg^-1 C^-1)

This amount of heat can be provided by condensing of steam=(4.186xx10^5)/(2.26 × 10^6)=0.185\ kg

Total steam condensed =0.151+0.185=0.336\ kg

Hence in the final mixture we will have
water at 100^@C=1.336\ kg and steam 99.664\ kg of steam