how much heat is required to vaporize 343g of liquid ethanol at its boiling point?
q = mass ethanol + heat vaporiation
288 j
To calculate the heat required to vaporize a liquid, you can use the formula:
Q = m * ∆Hv
Where:
Q is the heat required (in joules)
m is the mass of the substance (in grams)
∆Hv is the heat of vaporization (in J/g)
For ethanol, the heat of vaporization (∆Hv) is approximately 38.56 J/g.
Now, you can calculate the heat required to vaporize 343g of liquid ethanol:
Q = 343g * 38.56 J/g
Q = 13236.08 J
Therefore, it would require approximately 13,236.08 joules of heat to vaporize 343g of liquid ethanol at its boiling point.
To calculate the amount of heat required to vaporize a substance, you need to use the formula:
Q = m × ΔHvap
Where:
Q is the amount of heat required (in joules),
m is the mass of the substance (in grams), and
ΔHvap is the molar heat of vaporization (in J/g).
First, we need to determine the molar heat of vaporization for ethanol. The molar heat of vaporization, ΔHvap, is the amount of heat required to vaporize one mole of a substance. For ethanol, the molar heat of vaporization is around 38.6 kJ/mol, or 38.6 x 10^3 J/mol.
Next, we need to convert the mass of ethanol to moles. We can do this by using its molar mass. The molar mass of ethanol (C2H5OH) is roughly 46 g/mol.
Number of moles (n) = mass / molar mass
n = 343 g / 46 g/mol
n ≈ 7.46 mol
Now, we can calculate the amount of heat required:
Q = m × ΔHvap
Q = 343 g × (38.6 x 10^3 J/mol)
Q ≈ 13.24 x 10^6 J
Thus, approximately 13.24 x 10^6 joules of heat is required to vaporize 343 grams of liquid ethanol at its boiling point.