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3 May, 00:48

How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.

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  1. 3 May, 00:58
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    V H2O = 170.270 mL

    Explanation:

    QH2O (heat gained) = Qcoffe (heat ceded)

    ⇒ Q = mCΔT

    ∴ m: mass (g)

    ∴ C: specific heat

    assuming:

    δ H2O = δ Coffe = 1.00 g/mL C H2O = C coffe = 4.186 J/°C. g ... from literature

    ⇒ Q coffe = (mcoffe) (C coffe) (60 - 95)

    ∴ m coffe = (180mL) (1.00 g/mL) = 180 g coffe

    ⇒ Q = (180g) (4.186 J/°C. g) (-35°C) = - 26371.8 J

    ⇒ Q H2O = 26371.8 J = (m) (4.186 J/°C. g) (60 - 23)

    ⇒ (26371.8 J) / (154.882 J/g) = m H2O

    ⇒ m H2O = 170.270 g

    ⇒ V H2O = (170.270 g) (mL/1.00g) = 170.270 mL
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