How many gram of propane must be burned to supply the heat required to vaporize 6.75 L of water at 298 K?

Question

The equation of combustion is given and so is the delta H at -2.22 kJ (assuming per mole).

I'm confused on how to even start this problem, particularly with the vaporization aspect of water.

Answer 1

We will first determine the total amount of heat energy needed to raise 6.75 L of water from 25ºC (298K) to the boiling point of 100ºC. Then determine the heat needed to boil the water at 100º (phase change). The add them to get the total heat required. Then we will determine how much propane is needed to generate this much heat energy upon being combusted.

Step 1 - heat needed to raise temp of water from 25º to 100º

q = mC∆T

q = heat = ?

m = mass = 6.75 L x 1000 ml / L x 1 g / ml = 6750 g (assuming a density of 1g/ml for water)

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 75º

q = (6750 g)(4.184 J/gº)(75º) = 2,118,150 J = 2118 kJ

Step 2 - heat needed to boil the water at 100ºC (phase change, not change in temperature)

q = m∆Hvap

q = heat

m = mass = 6750 g

∆H = heat of vaporization for water = 2260 J/g (tabular value constant)

q = (6750 g)(2260 J/g) = 15,255,000 J = 15255 kJ

TOTAL HEAT required = 2118 kJ + 15255 kJ = 17373 kJ

Step 3 - how much propane is needed to produce 17373 kJ of heat?

Each mole of propane provides 2.22 kJ (SEE BELOW*) of heat so we can find moles and then grams.

moles of propane needed = 17373 kJ x 1 mol / 2.22 kJ = 7826 moles of propane needed'

grams of propane needed = 7826 mols x 44.1 g / mol = 345126 g of propane needed

*NOTE: The literature value for heat of combustion for propane is -2222 kJ/mol and not 2.22 kJ/mol

A more likely correct answer would be as follows:'

mols propane needed = 17373 kJ x 1 mol / 2222 kJ = 7.82 mols propane needed

grams propane needed = 7.82 mols x 44.1 g / mol = 345 g propane needed

Answer 2

Hey, here's how I approached this problem:

1. Water has a constant heat of vaporization (deltaH_vap, you shouldn't have to memorize this, it's a constant and usually given), which is equal to 40.7kJ/mol. Using the equation Q=deltaH*n, with n being equal to mols of water (using 1 mol water=18.02g and 1L=1000g), gives us 15200 kJ of energy needed to vaporize water.
2. BUT, aside from vaporizing it, the propane first had to raise the temperature to water's boiling point (100 degrees C) from 25 degrees Celsius (298 degrees K). To find how much energy this will take, use Q=mcdeltaT, using 6750g as m, c as the specific heat of liquid water (4.18 J/g degreesC, a constant), and 75.0 as your delta T, because you need to raise the temperature from 25.0 to 100. Celsius. This will result in 2120000J needed to raise the water's temperature to a boiling point, assuming my sig figs are correct.
3. Now, add 15200kJ and 2120000J (watch your units! 2120000J is 2120 kJ). You should get 17320 kJ.
4. Since combustion is an exothermic reaction, delta H will be negative. This is where -2.22 kJ/mol comes in.
5. HOWEVER, I am a bit confused by the provided delta H for propane's combustion, because it is normally around -2220 kJ/mol, not 22.2 kJ/mol. Could this be an error, or perhaps I'm interpreting the question wrong? Or, perhaps delta H is in MJ (megajoules), not kJ?
6. To finish the question, you will divide your total Q (17320 kJ) needed to vaporize water by the delta H combustion (in kJ/gram) and receive an answer in mols. Then, convert that answer into grams (molar mass propane 44.097g/mol), and you're done! I got 344g propane, using 2220 kJ/mol as delta H.
7. Please let me know if this was helpful or if you spot any mathematical/sig fig errors! This is a tricky question and took me a few attempts. Please note that this process also assumes the combustion of propane is perfectly efficient. The constants I used may slightly vary from what is given (for example, your molar masses may only be listed to the hundredth), resulting in slight variance.