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5 March, 10:03

Calculate the amount of heat liberated (in kJ) from 345 g of mercury when it cools from 72.7°C to 12.0°C.

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  1. 5 March, 10:57
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    2.93181 Kj

    Explanation:

    Given dа ta:

    Mass of mercury = 345 g

    Initial temperature = 72.7°C

    Final temperature = 12.0°C

    Heat released = ?

    Solution:

    Specific heat capacity:

    It is the amount of heat required to raise the temperature of one gram of substance by one degree.

    Specific heat capacity of mercury = 0.14 j/g.°C

    Formula:

    Q = m. c. ΔT

    Q = amount of heat absorbed or released

    m = mass of given substance

    c = specific heat capacity of substance

    ΔT = change in temperature

    ΔT = Final temperature - initial temperature

    ΔT = 12.0°C - 72.7°C

    ΔT = - 60.7°C

    Q = m. c. ΔT

    Q = 345 g * 0.14 j/g.°C * - 60.7°C

    Q = - 2931.81 j

    Joule to Kj conversion:

    2931.81 j * 1 kj / 1000 j = 2.93181 Kj
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