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 × 10 5 J kg –1 and latent heat of vaporization of water = 2.26 × 10 6 J kg –1 .

Question

Philoid · Accepted Answer

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

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.

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.