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0 votes
37.3k views
in Physics by (92.0k points)

The temperature of 170g of water 50°C is lowed to 5°C by adding certain amount of ice to it. Find the mass if ice added. 

Given :

Specific heat capacity of water = 4200 kg−1 °C−1 and Specific latent heat of ice = 336000 J kg−1

2 Answers

+1 vote
by (80.9k points)
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Best answer

Given mass of water = 170 g = 0.17 kg

T1 = 50°C, T2 = 5°C,

specific heat capacity of water = 4200 J kg –1°C–1

Sp. Latent heat of ice = 336000 J kg–1

Let mass of ice added be x then we have heat lost by water = Heat used by ice.

⇒ water at 50°C to 5°C = ice at °C to water at °C + water at 0°C to 5°C

⇒  we should add 90 g of ice to water

0 votes
by (10.9k points)
Let mass of ice be m kg.

Heat lost by  0.17kg of water at 50°C is

= 0.17×4200×(50-5)J

Heat gained by m kg ice to melt into water of 0°C is = 336000m J

Heat required to raise the temperature of m kg water from 0°C to 5°C is = m×4200×5 J

By calorimetric princppri we can write

336000m+m×4200×5=0.17×4200×55

 => m = (42×17×55)/357000=0.11 kg

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