1. A chemistry student added 22.5 grams of aluminum at 85.00C to 115 grams of water at 23.00C in a perfect calorimeter. The final temperature of the aluminum-water mixture was 41.40C. Use the student’s data to calculate the specific heat of aluminum in joules/gram0C.
2. Calculate the final temperature that results from mixing 100.0 grams of water at 1000C with 106.0 grams of water at 24.80C
[mass Al x specific heat Al x (Tfinal - Tinitial)] + [mass H2O x specific heat H2O x (Tfinal - Tinitial)] = 0
[22.5 g Al x sp. h. Al x (41.40 - 85.00)] + [115 x sp.h. H2O x (41.40 - 23.00)] = 0
You will need to look up the specific heat (sp.h.) of water.
The second problem is done the same way.
To calculate the specific heat of aluminum in joules/gram°C in question 1, we can use the equation:
q = m * c * ΔT
where:
q = heat exchanged
m = mass of the substance (in this case, aluminum)
c = specific heat capacity of the substance (what we want to find)
ΔT = change in temperature = final temperature - initial temperature
Now let's break down the steps to get the answer:
Step 1: Calculate the heat exchanged by the aluminum.
The heat exchanged by the aluminum is given by the equation:
q_aluminum = m_aluminum * c_aluminum * ΔT
where:
m_aluminum = mass of aluminum = 22.5 grams
ΔT = change in temperature = final temperature - initial temperature
In this case:
ΔT = 41.40°C - 85.00°C
Step 2: Calculate the heat exchanged by the water.
The heat exchanged by the water is given by the equation:
q_water = m_water * c_water * ΔT
where:
m_water = mass of water = 115 grams
c_water = specific heat capacity of water (4.18 J/g°C) since it's not given
ΔT = change in temperature = final temperature - initial temperature
In this case:
ΔT = 41.40°C - 23.00°C
Step 3: Since the calorimeter is perfect, we assume that no heat is lost to the surroundings. Therefore, the heat lost by the aluminum is equal to the heat gained by the water:
q_aluminum = -q_water
Step 4: Substitute the values and solve for the specific heat capacity of aluminum, c_aluminum:
m_aluminum * c_aluminum * ΔT_aluminum = -m_water * c_water * ΔT_water
c_aluminum = (-m_water * c_water * ΔT_water) / (m_aluminum * ΔT_aluminum)
Now plug in the values and calculate:
c_aluminum = (-115 g * 4.18 J/g°C * (41.40°C - 23.00°C)) / (22.5 g * (41.40°C - 85.00°C))
To get the answer, evaluate the expression on the right-hand side of the equation.
For question 2, you can use the same approach as above to calculate the final temperature resulting from mixing two bodies of water:
q1 + q2 = 0
where:
q1 = m1 * c1 * ΔT1
q2 = m2 * c2 * ΔT2
By using this equation, you can solve for the final temperature.