How much ice at a temperature of -22.3 ∘C must be dropped into the water so that the final temperature of the system will be 30.0 ∘C ?How much ice at a temperature of -22.3 ∘C must be dropped into the water so that the final temperature of the system will be 30.0 ∘C ?

111.6 g or 0.112 kg

Explanation:

specific heat of liquid water = 4190 J/kg⋅K

specific heat of ice = 2100 J/kg⋅K

heat of fusion for water = 3.34*10^5 J/kg

You didn't state the mass of the beaker, so, I will be assuming that the mass is negligible.

Assuming that the mass of ice required is m kg

Then the heat gained by the ice to attain zero degree will be

= m * 22.3 * 2100

= 46830m J

The heat gained by the ice to melt

= m * 3.34*10⁵ J

= 334000m J

The heat gained by water at zero degree to warm up to 30° =

m * 4190 * 30 = 125700m J

Total heat gained = 506530m J

Note: You didn't state the mass of the water and it's temperature, so I will be assuming that the mass of water is 0.3 kg, and it's temperature was 75° C

The heat lost by hot water to cool up to 30°

= .3 * 4190 * (75 - 30)

= 1257 * 45

= 56565 J

Using the relation, Heat lost = heat gained

506530m = 56565

m = 56565 / 506530 kg

m = 0.112 kg or 111.6 g

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