Suppose that 1.00 g of rubbing alcohol (C3H8O) evaporates from a 54.0 g aluminum block.
If the aluminum block is initially at 25 C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 C.
45.4 kJ/mole for the heat of vaporization of rubbing alcohol
The specific heat of Al is 0.903 J/g*C
1) Convert the 1.00 g of C3H8O to moles. (60 g per mole)
2) Compute the energy required to vaporize 1/60 mole (45.4/50 = 0.757 kJ). Call it Q
3) Compute the drop in temperature of the Al block when that amount of heat is removed. (delta T = Q/(M C) )
To find the final temperature of the aluminum block after the evaporation of alcohol, we need to calculate the heat lost by the aluminum and equate it to the heat gained by the alcohol during vaporization.
Step 1: Calculate the heat lost by the aluminum block.
The formula to calculate the heat lost is q = m * c * ΔT, where q is the heat lost, m is the mass, c is the specific heat, and ΔT is the change in temperature.
q = 54.0 g * 0.903 J/g°C * (final temperature - 25°C)
Step 2: Calculate the heat gained by the alcohol during vaporization.
The heat gained during the vaporization of 1.00 g of alcohol can be calculated using the formula q = n * ΔH, where q is the heat gained, n is the number of moles, and ΔH is the heat of vaporization.
First, calculate the number of moles of rubbing alcohol:
n = mass / molar mass = 1.00 g / (3 * 12.01 g/mol + 8 * 1.01 g/mol + 16.00 g/mol)
Step 3: Equate the two heat values.
Since the heat lost by the aluminum is equal to the heat gained by the alcohol, we can set up the equation:
q (heat lost by aluminum) = q (heat gained by alcohol)
54.0 g * 0.903 J/g°C * (final temperature - 25°C) = n (moles) * ΔH (heat of vaporization)
Step 4: Solve for the final temperature.
Plug in the given values and calculate the final temperature:
54.0 g * 0.903 J/g°C * (final temperature - 25°C) = (1.00 g / (3 * 12.01 g/mol + 8 * 1.01 g/mol + 16.00 g/mol)) * 45.4 kJ/mol
Solving for the final temperature will give you the answer.
To find the final temperature of the aluminum block after the evaporation of the alcohol, we can use the principle of energy conservation. The heat lost by the aluminum block will be equal to the heat gained by the alcohol for vaporization.
1. Calculate the heat lost by the aluminum block:
The heat lost by the aluminum block can be calculated using the equation: Q = mcΔT, where Q is the heat lost, m is the mass of the aluminum block, c is the specific heat of aluminum, and ΔT is the change in temperature.
Given that the mass of the aluminum block is 54.0 g and the specific heat of aluminum is 0.903 J/g°C, and assuming the final temperature of the aluminum block is T (unknown), we can calculate the heat lost by the aluminum block as follows:
Q_aluminum = mcΔT
Q_aluminum = 54.0 g * 0.903 J/g°C * (T - 25°C)
2. Calculate the heat gained by the alcohol for vaporization:
The heat gained by the alcohol for vaporization can be calculated using the equation: Q = nΔHvap, where Q is the heat gained, n is the number of moles of alcohol evaporated, and ΔHvap is the molar heat of vaporization of the alcohol.
Given that the molar heat of vaporization of rubbing alcohol (C3H8O) is 45.4 kJ/mol, and assuming the mass of the rubbing alcohol evaporated is 1.00 g, we can calculate the number of moles of alcohol evaporated:
n = mass / molar mass
n = 1.00 g / (12.0 g/mol + 3*1.01 g/mol + 16.0 g/mol)
n = 0.0385 mol
Then, we can calculate the heat gained by the alcohol for vaporization:
Q_alcohol = nΔHvap
Q_alcohol = 0.0385 mol * 45.4 kJ/mol
3. Set up the energy conservation equation:
Since energy is conserved, the heat lost by the aluminum block (Q_aluminum) should be equal to the heat gained by the alcohol for vaporization (Q_alcohol). Set up the equation:
Q_aluminum = Q_alcohol
Substitute the calculated values into the equation:
54.0 g * 0.903 J/g°C * (T - 25°C) = 0.0385 mol * 45.4 kJ/mol
4. Solve for the final temperature (T):
Now, we solve the equation to find the final temperature (T) of the aluminum block.
54.0 g * 0.903 J/g°C * (T - 25°C) = 0.0385 mol * 45.4 kJ/mol
Convert kJ to J:
54.0 g * 0.903 J/g°C * (T - 25°C) = 0.0385 mol * 45.4 kJ/mol * 1000 J/1 kJ
Simplify:
54.0 g * 0.903 J/g°C * (T - 25°C) = 0.0385 mol * 45,400 J/mol
Cancel out the units:
54.0 * 0.903 * (T - 25) = 0.0385 * 45,400
Multiply:
48.861 * (T - 25) = 1753.9
Expand:
48.861T - 48.861 * 25 = 1753.9
Rearrange the equation:
48.861T = 1753.9 + 48.861 * 25
Simplify:
48.861T = 3050.525
Divide by 48.861:
T = 62.5°C
Therefore, the final temperature of the aluminum block after the evaporation of the alcohol is 62.5°C.