2 gram of ice at zero degree celsius is mixed with 1 gram of steam at hundred degree Celsius find the temperature of the mixture.and also why we are getting 3g of H20in this problem.please explain
2 gram of ice at zero degree celsius is mixed with 1 gram of steam at hundred degree Celsius find the temperature of the mixture.and also why we are getting 3g of H20in this problem.please explain
Answer : 100°C Mixture Temperature
I am also going to assume that you have knowledge of
- latent heat of fusion and vaporization
- specific heat capacity
- Specific heat capacity of ice, water and steam are different.
Latent heat of fusion require to change ice into water without changing temperature
Latent heat of vapourization require to change water into heat without changing temperature
Specific heat of water require to change temperature of water
Latent Heat of fusion to change ice into water
Heat required to melt 1 g of ice of 0∘C to water at 0∘C = 1 × 80 cal.
Heat required to melt 2 g of ice of 0∘C to water at 0∘C = 2 × 80 cal = 160 cal
Now we have 2gm of water at 0∘C
Heat required to raise temperature of 1 g of water from 0∘C to 100∘C = 1 × 1 × 100 = 100 cal
Heat required to raise temperature of 2 g of water from 0∘C to 100∘C = 2 × 1 × 100 = 200 cal
Nowwe have 2gm of water at 100∘C
Total heat required by 2 gm of ice to attain maximum temperature of 100∘ = 160 + 200 = 360 cal
To supply this heat steam will convert into water and release latent heat of vapourisation.
1 gm steam when converted into water at 100°C release heat = 540 cal
So 2/3 gm steam when converted into water at 100°C will release heat = 360 cal
Hence final temperature of the mixture will be 100°C
Total water present = 2+2/3 gm = 2.67 gm at (100°C)
Total steam present = 1/3 gm =0 .33 gm at (100°C)
I do not know how you got 3 gm of water
If initially 3 gm of ice will be present
Than heat require will be 540 cal and full 1gm steam will be converted into water and finally you will get total water present = 3gm
Answer : 100°C Mixture Temperature
I am also going to assume that you have knowledge of
- latent heat of fusion and vaporization
- specific heat capacity
- Specific heat capacity of ice, water and steam are different.
Latent heat of vapourization require to change water into heat without changing temperature
Specific heat of water require to change temperature of water
Latent Heat of fusion to change ice into water
Heat required to melt 1 g of ice of 0∘C to water at 0∘C = 1 × 80 cal.
Heat required to melt 2 g of ice of 0∘C to water at 0∘C = 2 × 80 cal = 160 cal
Now we have 2gm of water at 0∘C
Heat required to raise temperature of 1 g of water from 0∘C to 100∘C = 1 × 1 × 100 = 100 cal
Heat required to raise temperature of 2 g of water from 0∘C to 100∘C = 2 × 1 × 100 = 200 cal
Nowwe have 2gm of water at 100∘C
Total heat required by 2 gm of ice to attain maximum temperature of 100∘ = 160 + 200 = 360 cal
To supply this heat steam will convert into water and release latent heat of vapourisation.
1 gm steam when converted into water at 100°C release heat = 540 cal
So 2/3 gm steam when converted into water at 100°C will release heat = 360 cal
Hence final temperature of the mixture will be 100°C
Total water present = 2+2/3 gm = 2.67 gm at (100°C)
Total steam present = 1/3 gm =0 .33 gm at (100°C)
I do not know how you got 3 gm of water
If initially 3 gm of ice will be present
Than heat require will be 540 cal and full 1gm steam will be converted into water and finally you will get total water present = 3gm
