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8 June, 16:51

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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  1. 8 June, 17:00
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    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
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