how much heat is required to vaporize 343 g of liquid ethanol at its boiling point?
To calculate the amount of heat required to vaporize a liquid, you can use the equation:
q = m * ΔHvap
Where:
q is the amount of heat required (in joules)
m is the mass of the liquid (in grams)
ΔHvap is the molar heat of vaporization (in J/g)
The molar heat of vaporization for ethanol is approximately 38.56 J/g.
Let's calculate the heat required to vaporize 343 g of liquid ethanol:
q = 343 g * 38.56 J/g
q = 13211.28 J
Therefore, approximately 13211.28 Joules of heat are required to vaporize 343 g of liquid ethanol at its boiling point.
To answer this question, we need to use the concept of heat of vaporization, also known as the molar heat of vaporization.
The molar heat of vaporization (ΔHvap) is the amount of heat required to convert one mole of a substance from the liquid phase to the vapor phase at a constant temperature and pressure. It is typically expressed in units of joules per mole (J/mol).
To determine the amount of heat required to vaporize a specific quantity of a substance, we can use the equation:
Q = m × ΔHvap
Where:
Q is the heat required (in joules)
m is the mass of the substance (in grams)
ΔHvap is the molar heat of vaporization (in J/mol)
Now, let's calculate the heat required to vaporize 343 g of liquid ethanol.
1. Find the molar mass of ethanol:
The molar mass of ethanol (C₂H₅OH) is 46.07 g/mol. This can be calculated by adding the atomic masses of carbon (12.01 g/mol), hydrogen (1.01 g/mol), and oxygen (16.00 g/mol).
2. Calculate the number of moles:
Divide the mass of ethanol (343 g) by the molar mass of ethanol (46.07 g/mol):
343 g / 46.07 g/mol ≈ 7.4514 mol
3. Look up the molar heat of vaporization:
The molar heat of vaporization of ethanol is approximately 38.56 kJ/mol (or 38560 J/mol). This value can be obtained from reference sources or databases.
4. Calculate the heat required:
Multiply the number of moles (7.4514 mol) by the molar heat of vaporization (38.56 kJ/mol or 38560 J/mol) to find the heat required:
Q = 7.4514 mol × 38560 J/mol ≈ 286,871 J
Therefore, approximately 286,871 joules (or 286.8 kJ) of heat is required to vaporize 343 g of liquid ethanol at its boiling point.