First, we need to calculate the mass of water and ethanol in the mixture. Since the density of both water and ethanol is 1 g/mL, we can directly use the volume to find the mass:mass_water = 300 mL * 1 g/mL = 300 gmass_ethanol = 200 mL * 1 g/mL = 200 gNext, we need to calculate the heat released due to the change in temperature for both water and ethanol:T_water = final_temperature - initial_temperature_water = 22C - 20C = 2CT_ethanol = final_temperature - initial_temperature_ethanol = 22C - 25C = -3Cq_water = mass_water * specific_heat_water * T_water = 300 g * 4.18 J/gC * 2C = 2512 Jq_ethanol = mass_ethanol * specific_heat_ethanol * T_ethanol = 200 g * 2.44 J/gC * -3C = -1464 JNow, we need to calculate the heat released due to the mixing of water and ethanol. First, we need to find the number of moles of water and ethanol:molar_mass_water = 18.015 g/molmolar_mass_ethanol = 46.07 g/molmoles_water = mass_water / molar_mass_water = 300 g / 18.015 g/mol = 16.66 molmoles_ethanol = mass_ethanol / molar_mass_ethanol = 200 g / 46.07 g/mol = 4.34 molThe heat of mixing is given per mole of the mixture. We need to find the total moles in the mixture and then multiply it by the heat of mixing:total_moles = moles_water + moles_ethanol = 16.66 mol + 4.34 mol = 21 molq_mixing = total_moles * heat_of_mixing = 21 mol * -21 kJ/mol = -441 kJ = -441,000 JFinally, we can find the total heat released by adding the heat released due to the change in temperature and the heat released due to mixing:q_total = q_water + q_ethanol + q_mixing = 2512 J - 1464 J - 441,000 J = -440,952 JThe heat released when 300 mL of water at 20C is mixed with 200 mL of ethanol at 25C is approximately 440,952 J.