# 1 kg of ice at – 10°C is mixed with 4.4 kg of water at 30°C. The final temperature of mixture is: (specific heat of ice is 2100 J/kg/k) .

Arun
25750 Points
4 years ago
Mass of ice, m = 1kg at -10°C
mass of water, M = 4.4kg at 30°C
heat flows through water to ice. so, heat gained by ice and heat lost by water.
total heat gained by ice to melt = heat required by ice to reach the temperature 0°C + heat require to melt of ice
= ms∆T + mL
= 1 kg × 2100 J/Kg/°C × (0 -(-10)) + 1kg × 336000/Kg
= 21000 + 336000
= 357000 J
heat lost by water to reduce the temperature at 0°C = MS∆T'
= 4.4kg × 4200 × (30 - 0) = 554400> 357000
so, temperature of mixture must be greater than zero.
now, let's take, after melting ice reaches temperature A.
then, heat is required to change temperature A is (554400 - 357000) = 197400J
use formula, (m + M)S∆T = 197400
or, (1 + 4.4)× 4200 × (A - 0) = 197400
A = 8.7°C

Vikas TU
14149 Points
4 years ago
Dear student
Check out this short method.
Given that,
Mass of ice = 1 kg
Temperature of ice = 283 K
Mass of water = 4.4 kg
Temperature of water = 303 K
Specific heat of ice = 2100 J/kg-K
We know that,
Specific heat of water = 4185.5 J/kg-K
Now, the final temperature of mixture is :
Tf = miTiCpi + mwTwCpw /(miCpi + mwCpw)
Put al the values
Tf = 301k
Hope this helps