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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) . `
2 months 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) = 197400A = 8.7°C
```
2 months ago
```							Dear student Check out this short method.Given that,Mass of ice = 1 kgTemperature of ice = 283 KMass of water = 4.4 kgTemperature of water = 303 KSpecific heat of ice = 2100 J/kg-KWe know that,Specific heat of water = 4185.5 J/kg-KNow, the final temperature of mixture is :Tf = miTiCpi + mwTwCpw /(miCpi + mwCpw)Put al the values Tf = 301k Hope this helps
```
2 months ago
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