Consider the combustion of propane. Delta H is -2221 kJ. Assume all heat is from combustion of propane. What mass of propane must be burned in order to furnish this amount of energy assuming the heat transfer process is 60% efficient.
Well, if we're talking about combustion, things are bound to get "hot" in here! Now, to calculate the mass of propane required, we need to consider the efficiency first. Since the heat transfer process is 60% efficient, it means only 60% of the energy released during combustion is actually transferred. Don't worry, those missing joules probably went on vacation or something.
Now, let's put on our math hats and get to work. We know that the enthalpy change, ΔH, is -2221 kJ. But remember, this is the amount of energy released, so we need to adjust for the 60% efficiency. Let's call the mass of propane we need to find "x" for now.
The energy transferred is given by the equation: Energy transferred = Efficiency * Energy released.
So, Energy transferred = 0.60 * (-2221 kJ).
Now we need to convert kJ to J because our calculator is too lazy to deal with kilo units. So, -2221 kJ becomes -2221 * 10^3 J.
Finally, we can set up an equation using the equation: Energy transferred = Mass of propane * heat combustion value (ΔH).
0.60 * (-2221 * 10^3 J) = x * (-2221 kJ).
Now we can cancel out the ΔH on both sides by dividing it out:
0.60 * (-2221 * 10^3 J) / (-2221 kJ) = x.
Cancelling out units, we get:
0.60 * (-2221 * 10^3 J) / (-2221 * 10^3 J/g) = x.
Dividing and multiplying all those numbers on the right side:
0.60 * (10^3 J) / (10^3 J/g) = x.
And voila! We're left with:
0.60 g = x.
So, to furnish this sizzling amount of energy with a 60% efficient heat transfer process, we would need to burn a mass of 0.60 grams of propane. That's about as light as a feather in the world of propane!
To find the mass of propane required to furnish a certain amount of energy, we need to use the equation:
ΔH = q = mcΔT
Where:
ΔH = Enthalpy change (in this case, -2221 kJ)
q = Heat transferred (in this case, -2221 kJ)
m = Mass of propane
c = Specific heat capacity of propane
ΔT = Change in temperature (assume constant)
Since all the heat transfer is from the combustion process, we can assume that the heat transferred is equal to the energy released by burning the propane.
First, we need to calculate the energy released by the combustion of propane.
Assuming the combustion process is 100% efficient, we can equate the heat transfer to the energy released:
q = -2221 kJ
Now, we also need to account for the fact that the heat transfer process is only 60% efficient. Therefore, we divide the calculated energy by the efficiency factor (0.60):
Energy released = q / 0.60
Now we can calculate the mass of propane using the equation:
q = mcΔT
Rearranging the equation to solve for mass:
m = q / (cΔT)
The specific heat capacity of propane (c) is approximately 1.99 kJ/kg°C. Assuming a constant temperature change, ΔT can be any value we choose.
Let's use ΔT = 1°C:
ΔT = 1°C = 1 K
Substituting the given values:
m = (-2221 kJ / (1.99 kJ/kg°C * 1 K)) / 0.60
m ≈ 1851 kg
Therefore, approximately 1851 kg of propane must be burned to furnish -2221 kJ of energy, assuming the heat transfer process is 60% efficient.
To solve this problem, we need to calculate the amount of propane that needs to be burned in order to furnish the given amount of energy. We'll also consider the heat transfer process, which is stated to be 60% efficient.
Step 1: Determine the quantity of energy needed
Given: ΔH = -2221 kJ (negative sign indicates that energy is released)
Step 2: Calculate the quantity of energy needed after considering efficiency
Since the heat transfer process is 60% efficient, we need to find the actual energy required. This can be calculated using the formula:
Actual energy needed = (Energy needed) / (Efficiency)
Actual energy needed = (-2221 kJ) / (0.60)
Step 3: Convert energy to joules
Since the given energy is in kilojoules, we need to convert it to joules for the calculations. There are 1000 joules in 1 kilojoule. Hence,
Actual energy needed = (-2221 kJ / 0.60) * (1000 J / 1 kJ)
Step 4: Calculate the mass of propane burned
The energy released during the combustion of propane can be calculated using the equation:
ΔH = (Energy released) / (Mass of propane)
So, we rearrange the equation to solve for the mass of propane:
Mass of propane = (Energy released) / ΔH
Substituting the given values:
Mass of propane = (Actual energy needed) / ΔH
Step 5: Substitute the values and calculate
Substitute the values we have obtained into the equation for mass of propane:
Mass of propane = (Actual energy needed) / ΔH
Mass of propane = [(−2221 kJ / 0.60) * (1000 J / 1 kJ)] / (-2221 kJ)
Simplify the equation and calculate the result to get the mass of propane needed.
2221 kJ is dH. You lose 40%.
So 60% of what number is 2221. That must be about 3700 kJ which is what you need (that's just a close guess)
C3H8 + 5O2 ==>3O2 + 4H2O
molar mass C3H8 is 44 grams
You know 44 g C3H8 must be burned to produce about 2221 kJ.
So common sense says
44 g x (3700/2221) = ?
In other words, 44 g, if 100% would produce 2221 kJ BUT the process in the problem loses 40% and you must actually produce about 3700 kJ. So that takes more propane. The ratio 3700/2221 tells you how much more.