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30 July, 17:55

if you combine 390.0 mL of water at 25.00 °C and 110.0 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. 30 July, 18:17
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    T (final) = 40.4 °C

    Explanation:

    Given dа ta:

    Density of water = 1.00 g/mL

    Volume of water at 25°C = 390.0 mL

    Volume of water at 95°C = 110.0 mL

    Final temperature of mixture = ?

    Solution:

    Heat absorbed by warm water = heat lost by hot water

    q lost = - q lost ... (1)

    Formula:

    q = m. c.ΔT

    ΔT (warm) = T (final) - 25°C

    ΔT (hot) = T (final) - 95°C

    Mass of water:

    d = m/v

    1 g/mL = m / 390 mL

    390 g = m

    For hot water:

    d = m/v

    1 g/mL = m / 110.0 mL

    110.0 g = m

    Now we will put the values in equation 1.

    390 g. c. (T (final) - 25°C) = - 110.0 g. c. (T (final) - 95°C)

    390 T (final) - 9750 °C = - 110 T (final) + 10450 °C

    390 T (final) + 110 T (final) = 10450 °C + 9750 °C

    500 T (final) = 20200

    T (final) = 20200/500

    T (final) = 40.4 °C
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