A 22.2-g piece of Al (which has a molar heat capacity of 24.03 J/mol°C) is heated to 84.6°C and dropped into a calorimeter containing water (the specific heat capacity of water is 4.18 J/g°C) initially at 23.5°C. The final temperature of the water is 27.8°C. Calculate the mass of water in the calorimeter.
heat lost by Al + heat gained by water = 0
[mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for the only unknown in the problem (which is mass H2O)
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To calculate the mass of water in the calorimeter, we can use the following equation:
q_water = - q_aluminum
Where q_water is the heat absorbed by the water and q_aluminum is the heat released by the aluminum.
The formula for calculating the heat absorbed or released is:
q = m * c * ΔT
Where:
q is the heat absorbed or released
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
We can calculate the heat absorbed by the water and the heat released by the aluminum.
For the water:
q_water = m_water * c_water * ΔT_water
For the aluminum:
q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
Now, let's calculate the heat absorbed by the water:
q_water = m_water * c_water * ΔT_water
c_water = 4.18 J/g°C
ΔT_water = (27.8 - 23.5)°C
q_water = m_water * 4.18 J/g°C * (27.8 - 23.5)°C
Next, let's calculate the heat released by the aluminum:
q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
c_aluminum = 24.03 J/mol°C
ΔT_aluminum = (27.8 - 84.6)°C
q_aluminum = 22.2 g * (1 mol/27.0 g) * 24.03 J/mol°C * (27.8 - 84.6)°C
Since q_water and q_aluminum are equal, we can set up an equation:
m_water * 4.18 J/g°C * (27.8 - 23.5)°C = 22.2 g * (1 mol/27.0 g) * 24.03 J/mol°C * (27.8 - 84.6)°C
Now we can solve for m_water:
m_water = (22.2 g * (1 mol/27.0 g) * 24.03 J/mol°C * (27.8 - 84.6)°C) / (4.18 J/g°C * (27.8 - 23.5)°C)
By simplifying this equation, we can calculate the mass of water in the calorimeter.