a piece of ice of mass 40 g is added to 200 g of water at 50 degree C. calculate the final temperature of water when all the ice has melted. specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of fusion of ice = 336*103 J kg-1.
a piece of ice of mass 40 g is added to 200 g of water at 50 degree C. calculate the final temperature of water when all the ice has melted. specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of fusion of ice = 336*103 J kg-1.
Latent heat of fusion of ice is, L = 336000 J/kg
Specific heat capacity of water, s = 4200 J/(kg-K)
Heat gained by 40 g or 0.040 kg ice when it melts at 0 oC = (0.040)(336000) = 13440 J
Suppose the final temperature of the mixture is ‘t’.
Heat gained by 0.040 kg water at 0 oC in converting into water at t oC is
= (0.040)(4200)t
= 168t J
Heat lost by 200 g or 0.2 kg water as its temperature decreases from 50 oC to t oC is
= (0.2)(4200)(50 – t)
= 42000 – 840t
Now,
Total heat lost by 0.2 kg water = total heat gained by 0.04 g ice
=> 42000 – 840t = 13440 + 168t
=> 1008t = 28560
=> t = 28.3 oC
This is the final temperature of the mixture.
Latent heat of fusion of ice is, L = 336000 J/kg
Specific heat capacity of water, s = 4200 J/(kg-K)
Heat gained by 40 g or 0.040 kg ice when it melts at 0 oC = (0.040)(336000) = 13440 J
Suppose the final temperature of the mixture is ‘t’.
Heat gained by 0.040 kg water at 0 oC in converting into water at t oC is
= (0.040)(4200)t
= 168t J
Heat lost by 200 g or 0.2 kg water as its temperature decreases from 50 oC to t oC is
= (0.2)(4200)(50 – t)
= 42000 – 840t
Now,
Total heat lost by 0.2 kg water = total heat gained by 0.04 g ice
=> 42000 – 840t = 13440 + 168t
=> 1008t = 28560
=> t = 28.3 oC
This is the final temperature of the mixture.
