Question

800g of ice of 0 degree Celsius is mixed with 200g of water vapour of 100 degree Celsius. What will be final temperature of mixture?​

Answer 1

Explanation:

Given 800 g of ice of 0 degree Celsius is mixed with 200 g of water vapour of 100 degree Celsius. What will be final temperature of mixture?

• We know that latent heat of fusion of ice = 336 J/g
• Specific heat of water = 4.186 J/g degree C
• First 800 g of ice melts and temperature increases by gaining heat from 200 g of water.
• Let final temperature be T
• So energy gained by 800 g of ice = energy lost by 200 g water.
• 800 x 336 + 800 x 4.186 x (T – 0) = 200 x 4.186 x (100 – T)
• 268,800 + 3,348.8 T = 83,720 – 837.2 T
• 4186 T = - 1,85,080

So T =  - 44.21 degree Celsius

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