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22 June, 10:14

A mixture containing 21.4g of ice (at exactly 0 C) and 75.3g of water (at 55.3 C) is placed in an insulated container. Assuming no heat is lost to the surroundings, what is the final temperature of the mixture?

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  1. 22 June, 12:16
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    Energy conservation law = > heat lost by the hot water = heat gained by ice

    At the equilibrium the temperature of mixture is T.

    heat lost by the hot water mCsΔT = 75.3g * 1 cal / (g°C) * (55.3°C - T)

    heat gained by the ice = latent fusion heat at 0°C + heat absorbed to rise temperature from 0°C to the equilibrium temperature of the mixture, T

    latent fusion heat of ice per gram, at 0°C = 79.7 cal/g

    latent fusion heat of ice at 0°C = 21.4g * 79.7 cal/g

    Heat absorbed to rise the temperature until thermal equilibrium:mCsΔT = 21.4g*1cal (g°C) * (T-0°C)

    Total heat gained by ice = total heat lost by the hot water

    21.4g * 79.7 cal/g + 21.4g*1cal / (g°C) * (T) = 75.3g * 1 cal / (g°C) * (55.3°C - T)

    1705.58 + 21.4 T = 4164.09 - 75.3T

    21.4T + 75.3T = 4164.09 - 1705.58

    96.7T = 2458.51

    T = 2458.51 / 96.7

    T = 25.42 °C

    Answer: 25.4 °C
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