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9 October, 12:30

how many joules of heat are absorbed when the tempurature of a 13.9 g sample of CaCO3 (s) increases from 21.7C to 33.3 C? specific heat of calcium carbonate is. 82 J/g-K

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  1. 9 October, 12:34
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    Q = 132.22 j

    Explanation:

    Given dа ta:

    Mass of CaCO₃ = 13.9 g

    Initial temperature = 21.7 °C

    Final temperature = 33.3 °C

    Specific heat of CaCO₃ = 0.82 j/g. k

    Amount of heat absorbed = ?

    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 = 33.3°C - 21.7°C

    ΔT = 11.6°C

    Q = m. c. ΔT

    Q = 13.9 g. 0.82 j/g. k. 11.6°C

    Q = 132.22 j
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