540g of ice at 0 degree C mixed with 540g of water at 80 degree Celsius final temperature of mixture is
Answers & Comments
punitithaka
We have mixed 540 g of ice at 0°c with 540 g of water at 80°c.
Given that latent heat of fusion = 80 cal/g sp. heat capacity of water =1 cal/g/°c sp. heat capacity of ice =0.5 cal/g/°c
So first thing in this mixing process will occur, is heat transfer from water at 80°c to saturated ice at 0°c till thermodynamic equilibrium occurs. Since ice is saturated, so it will absorb heat in the form of latent heat to convert into water. Let final temp.( equilibrium) of mixture is = T°c So energy balance equation,
Latent heat to ice at 0°c to convert into saturated water at 0°c + sensible heating of this saturated water from 0°c to final temp T°c = cooling of hot water from 80°c to final temp T°c
=> (540*80) + {540*1*(T-0)} = {540*1*(80-T)}
=> 540*T = -540*T
=> 1080*T = 0
=> T = 0
So final temp of mixture is 0°c and mixture is saturated water.
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Answers & Comments
have mixed 540 g of ice at 0°c with 540 g of water at 80°c.
Given that
latent heat of fusion = 80 cal/g
sp. heat capacity of water =1 cal/g/°c
sp. heat capacity of ice =0.5 cal/g/°c
So first thing in this mixing process will occur, is heat transfer from water at 80°c to saturated ice at 0°c till thermodynamic equilibrium occurs.
Since ice is saturated, so it will absorb heat in the form of latent heat to convert into water.
Let final temp.( equilibrium) of mixture is = T°c
So energy balance equation,
Latent heat to ice at 0°c to convert into saturated water at 0°c + sensible heating of this saturated water from 0°c to final temp T°c = cooling of hot water from 80°c to final temp T°c
=> (540*80) + {540*1*(T-0)} = {540*1*(80-T)}
=> 540*T = -540*T
=> 1080*T = 0
=> T = 0
So final temp of mixture is 0°c and mixture is saturated water.
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