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How many joules of heat are lost by 3580 KG granite as it goes from 41.2°C to - 12.9°C

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  1. Today, 01:15
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    The granite loses 1.53 * 10⁸ J of heat.

    The formula for the heat (q) transferred is

    q = mCΔT

    I shall assume that the specific heat capacity of granite is 0.790 J·°C⁻¹g⁻¹.

    m = 3.580 * 10⁶ g; ΔT = T_f - T_i = - 12.9 °C - 41.2 °C = - 54.1 °C

    ∴ q = 3.580 * 10⁶ g * 0.790 J·°C⁻¹g⁻¹ * (-54.1 °C) = - 1.53 * 10⁸ J

    The granite block loses 1.53 * 10⁸ J.
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