29.8 ml of ethanol (density = 0.789 g/ ml) initially at 9.9 degree celsius is mixed with 33.8 mL of water (density = 1.0 g/mL) initially at 27.8 degree celcius in an insulated beaker. Assuming that no heat is lost, what is the final temperature of mixture?
How do I start this problem?
heat gained by ethanol + heat lost by water = 0
heat gained by ethanol = mass x specific heat x (Tfinal-Tinitial). Use density to convert mL to mass.
heat lost by water = mass x specific heat x (Tfinal-Tinitial). Use density to convert mL to mass.
Tfinal is the only unknown.
To solve this problem, you can use the principle of conservation of energy. The heat gained by the ethanol and water will be equal to the heat lost by the surroundings. The formula to calculate heat transfer is:
Q = m * c * ΔT
Where:
Q is the heat transfer (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g·°C)
ΔT is the change in temperature (in °C)
Step 1: Calculate the initial and final temperatures of both ethanol and water.
Initial temperature of ethanol (T1_ethanol) = 9.9 °C
Initial temperature of water (T1_water) = 27.8 °C
Let the final temperature of the mixture be T_final.
Step 2: Calculate the initial heat of the ethanol and water.
Q_ethanol = m_ethanol * c_ethanol * ΔT_ethanol
Q_water = m_water * c_water * ΔT_water
m_ethanol = volume_ethanol * density_ethanol
m_water = volume_water * density_water
Given:
Volume of ethanol (V_ethanol) = 29.8 mL
Density of ethanol (density_ethanol) = 0.789 g/mL
Volume of water (V_water) = 33.8 mL
Density of water (density_water) = 1.0 g/mL
Step 3: Calculate the final heat of the mixture.
Q_final = Q_ethanol + Q_water
Since the beaker is insulated and assuming that no heat is lost, Q_final = 0.
Step 4: Set up equations for the heat transfers of ethanol and water in terms of the differences in temperature.
Q_ethanol = m_ethanol * c_ethanol * (T_final - T1_ethanol)
Q_water = m_water * c_water * (T_final - T1_water)
Step 5: Plug in the known values and solve for T_final.
Q_final = 0 (since no heat is lost)
m_ethanol = V_ethanol * density_ethanol
m_water = V_water * density_water
Q_ethanol = V_ethanol * density_ethanol * c_ethanol * (T_final - T1_ethanol)
Q_water = V_water * density_water * c_water * (T_final - T1_water)
0 = V_ethanol * density_ethanol * c_ethanol * (T_final - T1_ethanol) + V_water * density_water * c_water * (T_final - T1_water)
Solve the equation for T_final.
Note: Specific heat capacity values for ethanol and water can be found in reference books or online databases.