A 60.0 sample of water is heated to its boiling point. How much heat (in ) is required to vaporize it? (Assume a density of 1.00 .)
136 kJ
delta H x mass = q
40.65 kJ/kg x 0.060 kg = 135.4 kJ which rounds to 135 kJ.
By the way, you SHOULD have written units. It's 60 GRAMS and 1.00 g/mL.
By itself, 60 (or 1.00) doesn't mean a thing.
To calculate the heat required to vaporize the water, we need to use the formula for heat:
Q = m * ΔHv
Where:
Q = heat (in joules)
m = mass of water (in grams)
ΔHv = heat of vaporization of water (in J/g)
Given:
Mass of water (m) = 60.0 g
Density of water = 1.00 g/mL
First, we need to convert the mass of water to volume using the density formula:
Density = mass / volume
1.00 g/mL = 60.0 g / Volume
Rearranging the equation, we find Volume = 60.0 g / 1.00 g/mL
Volume = 60.0 mL
Since 1 mL of water has a mass of 1.00 g, we know that 60.0 mL of water will have a mass of 60.0 g.
Now we can calculate the heat:
Q = m * ΔHv
Q = 60.0 g * ΔHv
However, we still need the heat of vaporization (ΔHv) of water. The heat of vaporization of water is approximately 40.7 J/g.
Substituting this value:
Q = 60.0 g * 40.7 J/g
By multiplying 60.0 g and 40.7 J/g, we can find the heat required to vaporize the sample of water.
To calculate the amount of heat required to vaporize the water, you need to use the heat equation:
Q = m * ΔHv
where:
Q is the amount of heat required (in Joules),
m is the mass of the water (in grams), and
ΔHv is the heat of vaporization of water (in J/g).
In this case, the mass of the water is given as 60.0 grams.
The heat of vaporization of water is approximately 2260 J/g.
Plugging in the given values into the equation:
Q = 60.0 g * 2260 J/g
Calculating the product:
Q ≈ 135,600 J
Therefore, approximately 135,600 Joules of heat is required to vaporize the 60.0 gram sample of water.