A 13.0-g sample of ice at -13.0°C is mixed with 112.0 g of water at 80.0°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.
(heat to move ice from -13 to 0) + heat to melt ice) + (heat to move melted ice from 0 to final) + (heat lost by 112 g H2O @ 80) = 0
[(mass ice x specific heat ice x (0-(-13))] + [mass ice x heat fusio] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + (mass water x specific heat water x (Tfinal-Tinitial)] = 0
Solve for Tfinal. Answer is close to 63 C.
To calculate the final temperature of the mixture, we need to use the principle of conservation of energy. The heat lost by the water as it cools down (q1) must be equal to the heat gained by the ice as it warms up (q2).
First, let's calculate q1, the heat lost by the water.
q1 = mass_water × specific_heat_water × ΔT1
Where:
mass_water = 112.0 g (mass of water)
specific_heat_water = 4.18 J/g · °C (specific heat capacity of water)
ΔT1 = initial temperature - final temperature of water
ΔT1 = 80.0°C - final temperature
Now, let's calculate q2, the heat gained by the ice.
q2 = mass_ice × enthalpy_fusion + mass_ice × specific_heat_ice × ΔT2
Where:
mass_ice = 13.0 g (mass of ice)
enthalpy_fusion = 6.02 kJ/mol = 6.02 × 10^3 J/mol (enthalpy of fusion for ice)
specific_heat_ice = 2.08 J/g · °C (specific heat capacity of ice)
ΔT2 = final temperature - (-13.0°C) = final temperature + 13.0°C
According to the principle of conservation of energy, q1 = q2:
mass_water × specific_heat_water × ΔT1 = mass_ice × enthalpy_fusion + mass_ice × specific_heat_ice × ΔT2
Now we can substitute the known values into the equation and solve for the final temperature.
To solve this problem, we can use the concept of heat transfer and the principle of conservation of energy.
First, let's calculate the heat transferred from the water to the ice using the equation:
Q = m * C * ΔT
Where:
Q = heat transferred (in joules)
m = mass of the substance (in grams)
C = specific heat capacity (in J/g · °C)
ΔT = change in temperature (in °C)
For the water:
m_water = 112.0 g
C_water = 4.18 J/g · °C
ΔT_water = final temperature - initial temperature = final temperature - 80.0°C (since the initial temperature of water was 80.0°C)
So the heat transferred from the water to the ice is:
Q_water = m_water * C_water * ΔT_water
Now let's calculate the heat absorbed by the ice during the phase change from solid to liquid (melting). The enthalpy of fusion (ΔH_fusion) for ice is given as 6.02 kJ/mol.
To calculate the heat absorbed during the phase change, we need to convert the mass of ice to moles, then use the equation:
Q_fusion = n * ΔH_fusion
Where:
Q_fusion = heat absorbed during fusion (in joules)
n = number of moles of ice
ΔH_fusion = enthalpy of fusion (in J/mol)
First, let's calculate the number of moles of ice:
m_ice = 13.0 g
Molar mass of ice (H2O) = 18.015 g/mol
n_ice = m_ice / Molar mass of ice
Now we can calculate the heat absorbed during the phase change:
Q_fusion = n_ice * ΔH_fusion
Since the enthalpy of fusion is given in kJ/mol, we need to convert it to joules:
ΔH_fusion = 6.02 kJ/mol * 1000 J/kJ
Now, we can calculate the total heat transferred:
Q_total = Q_water + Q_fusion
Finally, with the total heat transferred, we can calculate the final temperature of the mixture:
Q_total = (m_water + m_ice) * C_final * ΔT_final
Rearranging the equation, we can solve for ΔT_final and then calculate the final temperature:
ΔT_final = Q_total / [(m_water + m_ice) * C_final]
Final temperature = initial temperature + ΔT_final
Substituting the given values and calculating the final temperature will give us the answer.