how much heat is required to vaporize 34.8g of water at 100 degrees celcious?
Note the correct spelling of celsius.
q = mass H2O x heat vaporization.
Q=4180×2.26×10^6
To determine the amount of heat required to vaporize water, we need to use the formula:
q = m * ΔHvap
where:
q is the heat required (in joules)
m is the mass of the substance (in grams)
ΔHvap is the heat of vaporization (in J/g)
The heat of vaporization for water is approximately 40.7 J/g.
So, using the given values:
m = 34.8g
ΔHvap = 40.7 J/g
Now, let's calculate the heat required:
q = m * ΔHvap
q = 34.8g * 40.7 J/g
q = 1419.36 J
Therefore, approximately 1419.36 joules of heat are required to vaporize 34.8 grams of water at 100 degrees Celsius.