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6 July, 05:45

If 41grams of water at 24°C absorbs 2208 J of heat energy, what will be the final temperature of the water?

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  1. 6 July, 08:34
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    The problem can solved using the heat equation which is expressed as:

    H = mCΔT

    where H is the energy absorbed or released, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.

    2208 J = 41 g x 4.18 J/g·°C x (T - 24 °C)

    T = 36.88 °C
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