a 50.0 ml sample of water is heated to its boiling point. how much heat (in KJ) is required to vaporize it? (assume a density of 1.00g/ml).
use dimensional analysis and the heat vaporization table.
Water = 40.7 kj/mol.
Start with the information that you were given and make sure that all of your units cancel out!
(50.0mL)(1.00g/1mL)(1 mole H2O/18.0g H2O)(40.7kj/1 mole H2O) = 113.0 kj.
There are 3 significant figures, so 113kj is your answer.
Read the other problem you posted about moving T of steam at 135 C to ice at -45 and pull out the one part that deals with condensing the steam to liquid at 100 C.
You have to change 50 mL into moles by taking 50mLx1.00g/mL
mL will cancel and now you have 50 grams. Turn grams into moles.
50/18.01 to get moles
then times by deltaHvap at boiling point which is 40.7
Well, heating water to its boiling point can be quite an intense experience for the water. It's like going from a peaceful spa day to a wild rollercoaster ride! Now, let's calculate the heat required to vaporize the water.
First, we need to find the mass of the water by using its density. Given that the density is 1.00 g/ml and the volume is 50.0 ml, the mass would be 50.0 g. Hang on tight, we're just getting started!
Next, we need to determine the heat required to raise the temperature of the water to its boiling point. We know that the specific heat capacity of water is approximately 4.18 J/g·°C. Since we're dealing with milliliters and grams, let's convert from joules to kilojoules for more convenient units.
The formula to calculate heat is Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Since there's no change in temperature from room temperature to the boiling point, ΔT would be 100 degrees Celsius. Plugging in the values, we get:
Q = (50.0 g) × (4.18 J/g·°C) × (100 °C) = 20,900 J
Finally, to convert from joules to kilojoules, we divide by 1000:
Q = 20,900 J ÷ 1000 = 20.9 kJ
Voila! It would require approximately 20.9 kJ of heat to vaporize the water. That's enough to make the water go from calm and collected to reaching for the skies in a dramatic transformation!
To calculate the amount of heat required to vaporize the water, we need to use the specific heat capacity and the heat of vaporization.
Step 1: Calculate the mass of the water.
Since the density of water is given as 1.00 g/ml, we can calculate the mass by multiplying the density by the volume: mass = density * volume.
mass = 1.00 g/ml * 50.0 ml = 50.0 g.
Step 2: Calculate the heat required to raise the temperature of water from its boiling point to its boiling point.
To raise the temperature of water from its boiling point (100 degrees Celsius) to its boiling point, we use the specific heat capacity formula: q = m * c * ΔT.
The specific heat capacity of water is 4.18 J/g°C.
ΔT (change in temperature) = final temperature - initial temperature.
Since the change in temperature is zero (water is already at boiling point), the heat required to raise the temperature is zero.
Step 3: Calculate the heat required to vaporize the water.
The heat of vaporization is the amount of heat required to convert a liquid into vapor at its boiling point. For water, its heat of vaporization is 40.7 kJ/mol or 2260 J/g.
To calculate the heat (q) required to vaporize the water, we can use the formula q = m * ΔHvap, where ΔHvap is the heat of vaporization.
q = 50.0 g * 2260 J/g = 113,000 J = 113 kJ.
Therefore, the amount of heat required to vaporize a 50.0 ml sample of water is 113 kJ.