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Mixing ice and water

  • Thread starter Jahnavi
  • Start date
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1. The problem statement, all variables and given/known data

1Kg ice at -10° C is mixed with 1 Kg water at 100°C .Then find equilibrium temperature.

2. Relevant equations

Specific heat of water = 1Cal/g/°C
Specific heat of ice = 0.5 Cal/g/°C
Latent heat of fusion of ice = 80 Cal/g
3. The attempt at a solution

Heat required by ice to reach at 0°C= 1000x0.5x10
= 5000Cal

Heat required by ice at 0°C to convert in water = 1000x80
= 80000Cal

Heat released by water when it moves from 100°C to 0°C = 1000x1x100= 100000Cal

Surplus heat = 15000 Cal

This surplus heat increases the temp. of 2Kg water at 0°C

15000Cal = 2000x1x∆t

Increase in temperature = 15000/2000

= 7.5°C

Is my answer correct ?
 

haruspex

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1. The problem statement, all variables and given/known data

1Kg ice at -10° C is mixed with 1 Kg water at 100°C .Then find equilibrium temperature.

2. Relevant equations

Specific heat of water = 1Cal/g/°C
Specific heat of ice = 0.5 Cal/g/°C
Latent heat of fusion of ice = 80 Cal/g
3. The attempt at a solution

Heat required by ice to reach at 0°C= 1000x0.5x10
= 5000Cal

Heat required by ice at 0°C to convert in water = 1000x80
= 80000Cal

Heat released by water when it moves from 100°C to 0°C = 1000x1x100= 100000Cal

Surplus heat = 15000 Cal

This surplus heat increases the temp. of 2Kg water at 0°C

15000Cal = 2000x1x∆t

Increase in temperature = 15000/2000

= 7.5°C

Is my answer correct ?
Looks good.
 

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