25.0 mL of ethanol (density = 0.789 g/mL) initially at 4.2°C is mixed with 35.7 mL of water (density = 1.0 g/mL) initially at 25.3°C in an insulated beaker. Assuming that no heat is lost, what is the final temperature of the mixture?
[(mass EtOH x specific heat EtOH x (Tfinal-Tinitial)] + [(mass water x specific heat water x (Tfinal-Tinitial)] = 0
Find mass EtOH and mass water from the density.
mass = volume x density. Tf is the only unknown.
To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat lost by one substance is equal to the heat gained by the other substance.
First, we need to calculate the heat lost or gained by each substance.
For the ethanol:
The specific heat capacity of ethanol is 2.44 J/g·°C.
The mass of ethanol can be calculated using the density and volume:
mass_ethanol = density_ethanol × volume_ethanol
mass_ethanol = 0.789 g/mL × 25.0 mL
mass_ethanol = 19.725 g
The change in temperature of ethanol can be calculated using the equation:
ΔQ_ethanol = mass_ethanol × specific heat capacity_ethanol × ΔT_ethanol
For the water:
The specific heat capacity of water is 4.18 J/g·°C.
The mass of water can be calculated using the density and volume:
mass_water = density_water × volume_water
mass_water = 1.0 g/mL × 35.7 mL
mass_water = 35.7 g
The change in temperature of water can be calculated using the equation:
ΔQ_water = mass_water × specific heat capacity_water × ΔT_water
Now, since the heat lost by ethanol is equal to the heat gained by water, we can set up the equation:
ΔQ_ethanol = ΔQ_water
Substituting the equations for heat for each substance and rearranging, we get:
mass_ethanol × specific heat capacity_ethanol × ΔT_ethanol = mass_water × specific heat capacity_water × ΔT_water
Rearranging and solving for ΔT_water, we get:
ΔT_water = (mass_ethanol × specific heat capacity_ethanol × ΔT_ethanol) / (mass_water × specific heat capacity_water)
Substituting the given values:
ΔT_water = (19.725 g × 2.44 J/g·°C × (T_final - 4.2°C)) / (35.7 g × 4.18 J/g·°C)
Simplifying the equation, we can solve for T_final:
T_final = ((19.725 g × 2.44 J/g·°C × ΔT_ethanol) / (35.7 g × 4.18 J/g·°C)) + 4.2°C
Now, substituting the given values for ΔT_ethanol (which is T_initial_water - T_final) and solving for T_final:
T_final = ((19.725 g × 2.44 J/g·°C × (25.3°C - T_final)) / (35.7 g × 4.18 J/g·°C)) + 4.2°C
Simplifying and solving the equation, we can find the final temperature of the mixture.