An insulated beaker with negligible mass contains liquid water with a mass of 0.240 kg and a temperature of 75.8°C. How much ice at a temperature of - 14.1°C must be dropped into the water so that the final temperature of the system will be 27.0°C? Take the specific heat of liquid water to be 4190 J/kg⋅K, the specific heat of ice to be 2100 J/kg⋅K, and the heat of fusion for water to be 3.34*105 J/kg.

Question

Physics

Jaydin Burke

1 December, 21:29

An insulated beaker with negligible mass contains liquid water with a mass of 0.240 kg and a temperature of 75.8°C. How much ice at a temperature of - 14.1°C must be dropped into the water so that the final temperature of the system will be 27.0°C? Take the specific heat of liquid water to be 4190 J/kg⋅K, the specific heat of ice to be 2100 J/kg⋅K, and the heat of fusion for water to be 3.34*105 J/kg.

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Ireland Sanchez · Answer 1

Let mass of ice required is m kg

Heat gained by ice to attain zero degree

= m x 14.1 x 2100

= 29610m J

Heat gained by ice to melt = m x 3.34 x 10⁵ J

= 334000m J

Heat gained by water at zero degree to warm up to 27 degree

m x 4190 x 27 = 113130 J

Total heat gained = 476740m J

Heat lost by hot water to cool up to 27 degree

.24 x 4190 x (75.8 - 27)

= 49073.28 J

Heat lost = heat gained

476740 m = 49073.28

m = 49073.28/476740 kg

=102.93 gm