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19 June, 14:55

How much ice at a temperature of - 17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 ∘C?

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  1. 19 June, 16:22
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    An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of - 17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg. K, the specific heat for ice to be 2100J/Kg. K, and the heat of fusion for water to be 334000J/kg.

    The answer to the above question is

    Therefore 0.1133 kg ice at a temperature of - 17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C

    Explanation:

    To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows

    ΔH = m*c*ΔT

    = 0.205*4190 * (79.9 - 31.0) = 42002.655 J

    Therefore fore the ice, we have

    Total heat = mi*L + mi*ci*ΔTi = mi*334000 + mi * 2100 * (0 - -17.5) = 42002.655 J

    370750*mi = 42002.655 J

    or mi = 0.1133 kg

    Therefore 0.1133 kg ice at a temperature of - 17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
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