Calculating Equilibrium Temperature of Ice and Water Mixture

[Jahnavi]

Homework Statement

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

Homework 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

The Attempt at a Solution

[/B]
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]

Homework Statement

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

Homework 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

The Attempt at a Solution

[/B]
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.