A 18.3 g piece of aluminum (which has a molar heat capacity of 24.03 J/oC mol) is heated to
82.4oC and dropped into a calorimeter containing water (specific heat capacity of water is
4.18 J/goC) initially at 22.3oC. The final temperature of the water is 24.9oC. Calculate the
mass of water in the calorimeter. Ignore significant figures for this problem.
Well, I must say, this question is quite "heating"! So, let's get to solving it.
To calculate the mass of water in the calorimeter, we need to use the formula:
m1c1ΔT1 = m2c2ΔT2
Here's how we'll plug in the values:
m1 = mass of aluminum = 18.3 g
c1 = molar heat capacity of aluminum = 24.03 J/oC mol
ΔT1 = change in temperature of aluminum = (final temperature of aluminum - initial temperature of aluminum) = (82.4oC - 22.3oC)
m2 = mass of water
c2 = specific heat capacity of water = 4.18 J/goC
ΔT2 = change in temperature of water = (final temperature of water - initial temperature of water) = (24.9oC - 22.3oC)
Now, let's rearrange the formula to solve for m2 (mass of water):
m2 = (m1c1ΔT1) / (c2ΔT2)
Substituting the values, we get:
m2 = (18.3 g * 24.03 J/oC mol * (82.4oC - 22.3oC)) / (4.18 J/goC * (24.9oC - 22.3oC))
Now, let me calculate that for you... *drumroll*
*munches on a clown-sized calculator*
And the mass of water in the calorimeter is approximately... *more munching* 262.5 grams!
So, you've got 262.5 grams of water in the calorimeter. Cheers to some "cool" calculations!
To solve this problem, we need to use the principle of conservation of energy. The heat gained by the water is equal to the heat lost by the aluminum.
First, let's calculate the heat lost by the aluminum:
Q_aluminum = m_aluminum * C_aluminum * ΔT_aluminum
Where:
m_aluminum is the mass of aluminum (given: 18.3 g)
C_aluminum is the molar heat capacity of aluminum (given: 24.03 J/oC mol)
ΔT_aluminum is the change in temperature for the aluminum (final temperature - initial temperature)
ΔT_aluminum = 82.4oC - 24.9oC = 57.5oC
Q_aluminum = 18.3 g * 1 mol/27 g * 24.03 J/oC mol * 57.5oC
Next, let's calculate the heat gained by the water:
Q_water = m_water * C_water * ΔT_water
Where:
m_water is the mass of water (unknown)
C_water is the specific heat capacity of water (given: 4.18 J/goC)
ΔT_water is the change in temperature for the water (final temperature - initial temperature)
ΔT_water = 24.9oC - 22.3oC = 2.6oC
Q_water = m_water * 4.18 J/goC * 2.6oC
Since the heat lost by the aluminum is equal to the heat gained by the water:
Q_aluminum = Q_water
18.3 g * 1 mol/27 g * 24.03 J/oC mol * 57.5oC = m_water * 4.18 J/goC * 2.6oC
Simplifying the equation:
18.3 g * 24.03 J/oC mol * 57.5oC = m_water * 4.18 J/goC * 2.6oC
Now, we can solve for the mass of water (m_water):
m_water = (18.3 g * 24.03 J/oC mol * 57.5oC) / (4.18 J/goC * 2.6oC)
m_water ≈ 561.7 g
Therefore, the mass of water in the calorimeter is approximately 561.7 g.
To solve this problem, we need to use the principle of conservation of energy. The heat lost by the aluminum (mcΔT) will be equal to the heat gained by the water (mcΔT). Here's how to calculate the mass of water in the calorimeter:
1. Calculate the heat lost by the aluminum:
Q_aluminum = mcΔT
where
m is the mass of the aluminum (18.3 g),
c is the molar heat capacity of aluminum (24.03 J/oC mol),
and ΔT is the change in temperature of the aluminum (T_final - T_initial).
ΔT = T_final - T_initial
ΔT = 82.4oC - 24.9oC = 57.5oC
Q_aluminum = (18.3 g) * (24.03 J/oC mol) * (57.5oC)
2. Calculate the heat gained by the water:
Q_water = mcΔT
where
m is the mass of the water,
c is the specific heat capacity of water (4.18 J/goC),
and ΔT is the change in temperature of the water (T_final - T_initial).
ΔT = T_final - T_initial
ΔT = 24.9oC - 22.3oC = 2.6oC
Q_water = m * (4.18 J/goC) * (2.6oC)
3. Since the heat lost by the aluminum is equal to the heat gained by the water, we can set up an equation:
Q_aluminum = Q_water
(18.3 g) * (24.03 J/oC mol) * (57.5oC) = m * (4.18 J/goC) * (2.6oC)
4. Solve for the mass of water, m:
m = (18.3 g) * (24.03 J/oC mol) * (57.5oC) / ((4.18 J/goC) * (2.6oC))
Calculate the value to find the mass of water in grams.
Change 18.3 g Al to mols. mols = grams/atomic mass. Then heat lost by Al + heat gained by water = 0
[mols Al x specific heat Al x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for mass H2O after substitution.