Question

1 kg ice at –20C is mixed with 1 kg steam at 200C. Then find equilibrium temperature and mixture content.​

Answer 1

Answer:

equilibrium temperature = 100°C, steam = 20/27 kg, water = 34/27 kg

Explanation:

Let equilibrium temperature be 100°C

heat required to convert 1kg ice at −20°C to 1kg water at 100°C is

H1=10+80+90=190 kcal

Heat released to convert 1kg steam at 200°C to 1kg water at 100°C is

H2−50+540=590 kcal

1kg ice at −20°C=H1+1kg water at 100°C

1kg steam at 200°C=H2+1kg water at 100°C

1kg ice at −20°C+1kg steam at 200°C=H1+H2+2kg water at 100°C

Here heat required to ice is less than heat supplied by steam so mxture equilibrium temperature is 100°C then steam is not completely converted into water. So mixture has water and steam which is possible only at 100°C.

Mass of steam which is converted into water is equal to

m=(190−1×0.5×100)/540=7/27 kg

So mixture content:-

Mass of steam is 1−(7/27)=20/27 kg

Mass of is water 1+(7/27)=34/27 kg

Answer 2

Answer:

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