To calculate the enthalpy change, we first need to find the final temperature of the mixture. We can do this by setting up an energy balance equation, where the heat gained by the colder substance equals the heat lost by the warmer substance.Let's denote the final temperature as Tf. The mass of water m_water is 50 g since the density of water is 1 g/mL , and the mass of ethanol m_ethanol is 30 g since the density of ethanol is also 1 g/mL .The heat gained by water q_water can be calculated using the formula:q_water = m_water C_water Tf - T_water The heat lost by ethanol q_ethanol can be calculated using the formula:q_ethanol = m_ethanol C_ethanol T_ethanol - Tf Since the heat gained by water equals the heat lost by ethanol, we can set up the following equation:m_water C_water Tf - T_water = m_ethanol C_ethanol T_ethanol - Tf Now, we can plug in the values and solve for Tf:50 g 4.18 J g^-1 C^-1 Tf - 20C = 30 g 2.44 J g^-1 C^-1 30C - Tf 209 J/C Tf - 20C = 73.2 J/C 30C - Tf Now, we can solve for Tf:209Tf - 4180 = 73.2 30 - 73.2Tf209Tf + 73.2Tf = 4180 + 2196282.2Tf = 6376Tf 22.6CNow that we have the final temperature, we can calculate the enthalpy change for both water and ethanol.For water:H_water = q_water = m_water C_water Tf - T_water H_water = 50 g 4.18 J g^-1 C^-1 22.6C - 20C H_water = 50 g 4.18 J g^-1 C^-1 2.6CH_water 543.8 JFor ethanol:H_ethanol = q_ethanol = m_ethanol C_ethanol T_ethanol - Tf H_ethanol = 30 g 2.44 J g^-1 C^-1 30C - 22.6C H_ethanol = 30 g 2.44 J g^-1 C^-1 7.4CH_ethanol -543.8 JSince the enthalpy change for water is positive heat is gained and the enthalpy change for ethanol is negative heat is lost , the overall enthalpy change for the system is approximately zero. This is because the heat gained by water is equal to the heat lost by ethanol, and there is no net change in enthalpy for the system.