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15 July, 08:13

A sample of aluminum absorbs 9.86 J of heat, upon which the temperature increases from 23.2°C to 30.5°C. Since the specific heat capacity of aluminum is 0.90 J/g-K, what is the mass of the sample?

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  1. 15 July, 08:21
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    1.50 g

    Explanation:

    The heat absorbed by the aluminum in this case is:

    q = m x C x ΔT m = q / (C x ΔT)

    q = 9.86 J

    C = 0.90 J/g-K

    ΔT = (30.5 ºC - 23.2 ºC) = 7.3 ºC = 7.3 K (this is a range of temperature)

    m = 9.86 J / (0.90 J/g-K) x 7.3 K) = 1.50 g
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