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) .

Question

Arun · Answer

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 · Answer

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

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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) .

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) .

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