A 10.0 g sample of ice at -14.0°C is mixed with 124.0 g of water at 81.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J/g°C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
q1 = heat to raise T of ice from -14.0 C to zero C.
q1 = mass ice x heat capacity ice x (Tf-Ti) where Tf is 0 and Ti = -14.
q2 = heat to melt ice
q2 = mass iced x heat fusion.
q3 = heat to raise T of melted ice from zero C to Tfinal.
q3 = mass melted H2O x specific heat H2O x (Tf - Ti) where Tf is the unknown and Ti = 0
q4 = heat released by warm water at 81.0C dropping to Tfinal.
q1 + q2 + q3 + q4 = 0.
Substitute and solve for Tf. Post you work if you get stuck.
To calculate the final temperature of the mixture, you need to consider the heat gained or lost by both the ice and water as they undergo a temperature change.
First, let's calculate the heat gained or lost by the ice as it warms up from -14.0°C to its melting point at 0°C:
Q_ice = mass_ice × specific_heat_ice × ΔT_ice
Where:
- Q_ice is the heat gained or lost by the ice
- mass_ice is the mass of the ice (given as 10.0 g)
- specific_heat_ice is the specific heat capacity of ice (given as 2.08 J/g°C)
- ΔT_ice is the change in temperature of the ice (0°C - -14.0°C)
Plugging in the values:
Q_ice = 10.0 g × 2.08 J/g°C × (0 - (-14.0))°C
Next, let's calculate the heat gained or lost during the phase change as the ice melts into water:
Q_phase_change = moles_ice × enthalpy_fusion_ice
Where:
- Q_phase_change is the heat gained or lost during the phase change
- moles_ice is the moles of ice (to calculate this, divide the mass of ice by the molar mass of water, which is 18.015 g/mol)
- enthalpy_fusion_ice is the enthalpy of fusion for ice (given as 6.02 kJ/mol, which should be converted to J/mol)
Plugging in the values:
Q_phase_change = (10.0 g ÷ 18.015 g/mol) × 6.02 kJ/mol × 1000 J/kJ
Now, let's calculate the heat gained or lost by the water as it cools down from 81.0°C to the final temperature:
Q_water = mass_water × specific_heat_water × ΔT_water
Where:
- Q_water is the heat gained or lost by the water
- mass_water is the mass of the water (given as 124.0 g)
- specific_heat_water is the specific heat capacity of liquid water (given as 4.18 J/g°C)
- ΔT_water is the change in temperature of the water (final temperature - initial temperature)
Plugging in the values:
Q_water = 124.0 g × 4.18 J/g°C × (final_temperature - 81.0)°C
Since there is no heat loss to the surroundings, the heat gained by the ice and the heat gained by the water should be equal. Therefore, we can set up the equation:
Q_ice + Q_phase_change = Q_water
(10.0 g × 2.08 J/g°C × (0 - (-14.0))°C) + (10.0 g ÷ 18.015 g/mol) × 6.02 kJ/mol × 1000 J/kJ = 124.0 g × 4.18 J/g°C × (final_temperature - 81.0)°C
Simplifying and solving for the final temperature will give you the answer.