if you want to completey vaporize 6.0g H2O liquid initally at 48.0 degree celcuis how many kj would be needed
Q= cm Δt+Lm =4180•0.006•(100-48) + 2260000•0.006 = ….. (Joules)
To calculate the energy required to completely vaporize a substance, you need to use the heat of vaporization (ΔHvap) of the substance and its mass.
The heat of vaporization (ΔHvap) of water is 40.7 kJ/mol. We first need to convert grams into moles using the molar mass of water, which is approximately 18.015 g/mol.
Number of moles of water (n) = mass of H2O / molar mass of H2O
n = 6.0 g / 18.015 g/mol
n ≈ 0.333 mol
Next, we can calculate the energy required using the equation:
Energy (Q) = ΔHvap * n
Q = 40.7 kJ/mol * 0.333 mol
Q ≈ 13.53 kJ
Therefore, approximately 13.53 kJ of energy would be required to completely vaporize 6.0 grams of water initially at 48.0 degrees Celsius.