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22 July, 09:30

If 250.0 g of water at 30.0 °C cool to 5.0 °C, how many kilojoules of energy did the water lose?

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  1. 22 July, 13:00
    0
    -26.125 kj

    Explanation:

    Given dа ta:

    Mass of water = 250.0 g

    Initial temperature = 30.0°C

    Final temperature = 5.0°C

    Amount of energy lost = ?

    Solution:

    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 = T2 - T1

    ΔT = 5.0°C - 30.0°C

    ΔT = - 25°C

    Specific heat of water is 4.18 j/g.°C

    Now we will put the values in formula.

    Q = m. c. ΔT

    Q = 250.0 g * 4.18 j/g.°C * - 25°C

    Q = - 26125 j

    J to kJ

    -26125 j * 1 kj / 1000 j

    -26.125 kj
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