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27 April, 23:32

How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 163 mL of coffee at 97.9°C so that the resulting combination will have a temperature of 54.0°C? Assume that coffee and water have the same density and the same specific heat (4.18 J/g·°C) across the temperature range.

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  1. 28 April, 01:04
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    248 mL

    Explanation:

    According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.

    Qw + Qc = 0

    Qw = - Qc [1]

    We can calculate each heat using the following expression.

    Q = c * m * ΔT

    where,

    c: specific heat m: mass ΔT: change in the temperature

    163 mL of coffee with a density of 0.997 g/mL have a mass of:

    163 mL * 0.997 g/mL = 163 g

    From [1]

    Qw = - Qc

    cw * mw * ΔTw = - cc * mc * ΔTc

    mw * ΔTw = - mc * ΔTc

    mw * (54.0°C-25.0°C) = - 163 g * (54.0°C-97.9°C)

    mw * 29.0°C = 163 g * 43.9°C

    mw = 247 g

    The volume corresponding to 247 g of water is:

    247 g * (1 mL/0.997 g) = 248 mL
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