A 3.04 g piece of aluminum at 60.0 °C is placed into a “styrofoam cup” calorimeter filled with
150.0 mL of water at 25.0 °C. What is the final temperature of the contents of the calorimeter
once thermal equilibrium has been established (everything has the same final temperature)?
You may need to look up a heat capacity
Tfinal= m1c1T1 + m2c2T2/(m1c1+m2c2)
I found this equation to solve for Tfinal, but not sure how it was derived or if its the correct method to solve this problem
heat lost by Al + heat gained by H2O = 0
[mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
The equation you found may be right. I didn't go through the algebra.
To find the final temperature of the contents of the calorimeter, we need to consider the heat gained or lost by each substance involved.
First, let's calculate the heat gained or lost by the aluminum.
The heat gained or lost by a substance can be calculated using the equation:
Q = m * c * ΔT
Where:
Q is the heat gained or lost (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 (final temperature - initial temperature) of the substance (in °C)
Given:
Mass of aluminum (m) = 3.04 g
Specific heat capacity of aluminum = 0.897 J/g°C
Initial temperature of aluminum (Ti) = 60.0°C
Next, let's calculate the heat gained or lost by the water.
Using the same equation, we have:
Q = m * c * ΔT
Where:
Q is the heat gained or lost (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 (final temperature - initial temperature) of the substance (in °C)
Given:
Mass of water (m) = 150.0 mL = 150.0 g (since the density of water is approximately 1 g/mL)
Specific heat capacity of water = 4.184 J/g°C
Initial temperature of water (Ti) = 25.0°C
Now, let's assume the final temperature of the system is Tf.
Since we want to reach thermal equilibrium where everything has the same final temperature, the heat lost by the aluminum must equal the heat gained by the water. Therefore:
Q_aluminum = Q_water
Using the equations we derived earlier, and rearranging them, we have:
m_aluminum * c_aluminum * (Tf - Ti_aluminum) = m_water * c_water * (Tf - Ti_water)
Plugging in the given values:
(3.04 g) * (0.897 J/g°C) * (Tf - 60.0°C) = (150.0 g) * (4.184 J/g°C) * (Tf - 25.0°C)
Now we can solve for Tf by rearranging the equation and isolating Tf:
(3.04 g) * (0.897 J/g°C) * Tf - (3.04 g) * (0.897 J/g°C) * 60.0°C = (150.0 g) * (4.184 J/g°C) * Tf - (150.0 g) * (4.184 J/g°C) * 25.0°C
(3.04 g) * (0.897 J/g°C) * Tf - (150.0 g) * (4.184 J/g°C) * Tf = (150.0 g) * (4.184 J/g°C) * 25.0°C - (3.04 g) * (0.897 J/g°C) * 60.0°C
0.897 J/g°C * Tf * (3.04 g - 150.0 g) = (150.0 g) * (4.184 J/g°C) * 25.0°C - (3.04 g) * (0.897 J/g°C) * 60.0°C
Now we can solve for Tf:
Tf = [(150.0 g) * (4.184 J/g°C) * 25.0°C - (3.04 g) * (0.897 J/g°C) * 60.0°C] / (0.897 J/g°C * (3.04 g - 150.0 g))
Calculate the right side of the equation, and then divide it by the left side to find the final temperature Tf.
Please note that since the mass of the water is given in mL, we assume its density is approximately 1 g/mL, which is valid for water. In actual experiments, it's important to check the density of the specific substance.