If 55.0 mL of ethanol (density = 0.789 g/mL) initially at 9.0 °C is mixed with 55.0 mL of water (density = 1.0 g/mL) initially at 28.7 °C in an insulated beaker, what is the final temperature of the mixture, assuming that no heat is lost and the specific heat capacities are 2.44 J/g°C for ethanol and 4.18 J/g°C for water?

Set up the heat transfer equation based on the principle of conservation of energy, where the heat lost by water equals the heat gained by ethanol: \( m_{\text{water}} c_{\text{water}} (T_{\text{final}} - T_{\text{initial, water}}) = -m_{\text{ethanol}} c_{\text{ethanol}} (T_{\text{final}} - T_{\text{initial, ethanol}}) \).

Substitute the known values into the heat transfer equation: \( (55.0 \, \text{g}) (4.18 \, \text{J/g°C}) (T_{\text{final}} - 28.7 \, °C) = -(43.395 \, \text{g}) (2.44 \, \text{J/g°C}) (T_{\text{final}} - 9.0 \, °C) \).