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5 August, 22:26

Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C when it absorbs 3.42 kilojoules of energy.

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  1. 6 August, 00:09
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    Answer: 0.036 J/g°C

    Explanation:

    The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

    Thus, Q = MCΦ

    Given that,

    Q = 3.42 Kilojoules

    [Convert 3.42 kilojoules to joules

    If 1 kilojoule = 1000 joules

    3.42 kilojoules = 3.42 x 1000 = 3420J]

    Mass = 2.508Kg

    [Convert 2.508 kg to grams

    If 1 kg = 1000 grams

    2.508kg = 2.508 x 1000 = 2508g]

    C = ? (let unknown value be Z)

    Φ = (Final temperature - Initial temperature)

    = 42.061°C - 4.051°C

    = 38.01°C

    Apply the formula, Q = MCΦ

    3420J = 2508g x Z x 38.01°C

    3420J = 95329.08g•°C x Z

    Z = (3420J / 95329.08g•°C)

    Z = 0.03588 J/g°C

    Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

    Thus, the specific heat of the substance is 0.036 J/g°C
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