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```
How much ice at 0°C must be added to 100 g water at 30°C in order to reduce its temperature to 20°C? (The latent heat of fusion of water is 80 cal/g)
How much ice at 0°C must be added to 100 g water at 30°C in order to reduce its temperature to 20°C?
(The latent heat of fusion of water is 80 cal/g)

```
9 years ago

melvin davis
6 Points
```							heat lost by the water = heat gained by ice for changing into water and rising the temp. to 20degree celsius
100*1*(30-20) = m*80 + m*1*(20-0)
solve it
```
9 years ago
Sudheesh Singanamalla
114 Points
```							specific heat capacity of water = 1 cal / g ; Tf = 20 ; Ti = 30 ;
Heat lost by water = m * s * (Tf - Ti )
= 100 * 1 cal/g * ( 20 - 30 ) = -1000 J
Heat required to melt ice = m L = m * 80
Heat required to raise temperature of ice water to 20 = m * 1 cal/g * 10
heat lost = heat gained
-1000 = -m * 80 + 10 * m
-1000 = -90 * m
mass of ice (m) = 1000/90 = 100/9 = 11.11 grams.

Therefore 1.11 grams of ice is required to change water from 30 degrees to 20 degree celcius.
```
9 years ago
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• Test paper with Video Solution
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• Discussion Forum
• Previous Year Exam Questions