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4 October, 06:24

If you have 430.0 mL of water at 25.00 °C and add 140.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

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  1. 4 October, 06:53
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    Heat released by the hot water = mCeΔT = 140.0 g Ce (95.00°C - T)

    Heat absorbed by the cold water = mCeΔT = 430.0 g Ce (T - 25.00°C)

    Heat released = heat absorbed

    140.0*Ce (95 - T) = 430.0*Ce (T - 25)

    95 - T = 3.0714 (T - 25)

    95 - T = 3.0717T - 76.786

    3.071T + T = 95 + 76.786

    4.071T = 171.786

    T = 42.20 °C

    Answer: 42.20 °C
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