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15 January, 23:33

Calculate the amount of heat absorbed when 70 mL of water is heated from 30°C to 70°C. The specific heat of liquid water is 4.184 J/g °C. The density of water is about 1 g/mL. Report your answer in kJ to one decimal place.

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  1. 15 January, 23:58
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    the heat absorbed by water is Q=11715.2 J

    Explanation:

    the amount of heat absorbed is

    Q = m * c * (T - Ti)

    where

    Q = heat absorbed by water

    m = mass f water heated

    c = specific heat of water

    T = final temperature, Ti = inicial temperature

    also we know that the mass is related with the density through

    density = mass/volume

    d = m / V → m = d * V

    replacing m in the heat equation

    Q = m * c * (T - Ti) = d * V * c * (T - Ti)

    replacing values

    Q = d * V * c * (T - Ti) = 1 g/ml * 70 ml * 4.184 J/g°C * (70°C - 30°C) = 11715.2 J

    Note:

    we assume

    - constant density of the water between 30° and 70°

    - constant specific heat of water between 30° and 70°

    - the water has no impurities
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