Answer:
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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