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23 January, 06:16

A 34.34 g sample of a substance is initially at 26.7 °C. After absorbing 2205 J of heat, the temperature of the substance is 152.1 °C. What is the specific heat (c) of the substance?

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  1. 23 January, 08:12
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    The specific heat of the substance is c = 512.04 J/kg K

    Explanation:

    ΔQ = 2205 J

    m = 0.03434 kg

    ΔT = 125.4 ºC

    ΔQ = m * c * ΔT

    c = ΔQ / (m * ΔT)

    c = 512.04 J/Kg K
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