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17 May, 23:32

The specific heat capacity of gold is 0.128 J/g ºC. How much heat would be needed to warm 250.0 grams of gold from 25.0 ºC to 100.0 ºC? A. 2.4 x10-3J B. - 2.40 x 103J C. 2.40 x 103J D. - 2.40 x 10-3J

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  1. 18 May, 02:49
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    C. 2.40 * 10³ J

    Explanation:

    The formula for the amount of heat (q) absorbed is

    q = mCΔT

    1. Calculate ΔT

    ΔT = 100.0 °C - 25.0 °C = 75.0 °C

    2. Calculate q

    q₂ = mCΔT = 250.0 g * 0.128 J·°C⁻¹g⁻¹ * 75.0 °C = 2.40 * 10³ J
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