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17 November, 20:32

4130 J of energy is added to a 52 g sample of water that has an initial temperature of 10.0°C. If the specific heat of water is 4.18 J / (g * °C), what would the final temperature of the water be?

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  1. 17 November, 23:19
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    Q=mCΔT

    4130 = (52) (4.18) (Tf-10)

    4130 = (217.36) (Tf-10)

    (4130/217.36) + 10=Tf

    Tf=29°C
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