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.

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