LP gas burns according to the exothermic reaction:
C3H8(g) + 5 O2(g) right arrow 3 CO2(g) + 4 H2O(g) ΔHrxn° = −2044 kJ
What mass of LP gas is necessary to heat 1.2 L of water from room temperature (25.0°C) to boiling (100.0°C)? Assume that during heating, 18% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.
My calculator is broken; I'll try to describe this. First, since the LP gas is only 18% efficient, we may as well adjust that early.
2044 kJ(@100%) x 0.18 = about 400 kJ at 18%.
q needed = mass H2O x specific heat H2O x (Tfinal-Tinitial) = estimated 360,000 J.
So 44 g C3H8(1 mol) will produce about 400 kJ at 18% efficiency. What amount will produce 360 kJ?
44 g x (360/400) = ? g LP gas.
Well, let's start by calculating the amount of heat needed to raise the temperature of the water.
The specific heat capacity of water is 4.18 J/g·°C, so first, we need to convert the given volume of water into grams. Given that the density of water is approximately 1 g/mL, we have:
1.2 L × 1000 mL/L = 1200 g
Next, we need to find the temperature change:
ΔT = final temperature - initial temperature
ΔT = 100.0°C - 25.0°C
ΔT = 75.0°C
Now, let's calculate the heat needed to raise the temperature of the water:
q = m × ΔT × c
q = 1200 g × 75.0°C × 4.18 J/g·°C
q = 376,200 J
Since only 18% of the heat emitted by the LP gas is used to heat the water, we can calculate the total heat emitted by the LP gas:
Q_total = q / 0.18
Q_total = 376,200 J / 0.18
Q_total = 2,090,000 J
Now, let's convert the heat emitted by the LP gas into kilojoules:
Q_total = 2,090,000 J = 2090 kJ
According to the balanced exothermic reaction, the combustion of 1 mole of LP gas releases 2044 kJ of heat. So, to find the mass of LP gas needed, we can use a simple ratio:
2044 kJ of LP gas = 1 mole of LP gas
2090 kJ of LP gas = x mole of LP gas
x = (2090 kJ) / (2044 kJ/mole)
x ≈ 1.022 moles
Now, we need to convert moles of LP gas to grams. The molar mass of C3H8 is approximately 44.1 g/mol:
mass = 1.022 moles × 44.1 g/mol
mass ≈ 45 g
So, approximately 45 grams of LP gas is necessary to heat 1.2 L of water from room temperature to boiling. Just remember, when it comes to LP gas, always keep it light-hearted and not too hot!
To find the mass of LP gas needed to heat the water, we need to follow these steps:
Step 1: Calculate the heat required to raise the temperature of water from room temperature to boiling.
The specific heat capacity of water is approximately 4.18 J/g°C.
ΔT = (100.0°C - 25.0°C) = 75.0°C
q1 = mass × specific heat capacity × ΔT
Step 2: Calculate the heat absorbed by the water.
According to the problem statement, only 18% of the heat emitted by the LP gas combustion is used to heat the water. So we need to multiply the heat calculated in Step 1 by 0.18 to find the heat absorbed by the water.
q2 = q1 × 0.18
Step 3: Calculate the moles of LP gas burned.
From the balanced chemical equation, we know that 1 mole of C3H8 produces -2044 kJ of heat. We need to calculate the moles of LP gas burned to produce the heat absorbed by the water.
ΔHrxn = -2044 kJ
q2 = n × ΔHrxn
Step 4: Find the mass of LP gas.
From the given balanced chemical equation, we know that 1 mole of C3H8 has a molar mass of 44.1 g/mol (molar mass of C3H8 = 3 × atomic mass of C + 8 × atomic mass of H).
molar mass C3H8 = 3 × atomic mass C + 8 × atomic mass H
= (3 × 12.01 g/mol) + (8 × 1.01 g/mol)
= 44.1 g/mol
mass = n × molar mass
Now we will calculate the mass of LP gas needed to heat the water.
Let's substitute the given values:
Step 1:
ΔT = 75.0°C
q1 = (1.2 L × 1 g/mL) × 4.18 J/g°C × 75.0°C
Step 2:
q2 = q1 × 0.18
Step 3:
ΔHrxn = -2044 kJ
q2 = n × ΔHrxn
Step 4:
molar mass = 44.1 g/mol
mass = n × molar mass
Let's perform the calculations:
Step 1:
ΔT = 75.0°C
q1 = (1.2 L × 1 g/mL) × 4.18 J/g°C × 75.0°C
q1 = 376.2 J
Step 2:
q2 = q1 × 0.18
q2 = 376.2 J × 0.18
q2 = 67.88 J
Step 3:
ΔHrxn = -2044 kJ
q2 = n × ΔHrxn
67.88 J = n × -2044000 J/mol
n = -67.88 J / -2044000 J/mol
n = 3.32 × 10^-5 mol
Step 4:
molar mass = 44.1 g/mol
mass = n × molar mass
mass = (3.32 × 10^-5 mol) × (44.1 g/mol)
mass = 0.00146 g
Therefore, you would need approximately 0.00146 grams of LP gas to heat 1.2 L of water from room temperature to boiling.
To solve this problem, we need to follow a step-by-step approach. Here's how we can calculate the mass of LP gas required to heat the water:
Step 1: Determine the heat required to raise the temperature of water from 25.0°C to 100.0°C.
The specific heat capacity of water is approximately 4.184 J/g°C. So, we need to calculate the heat required to raise the temperature of water from 25.0°C to 100.0°C.
Q = m * c * ΔT
Where:
Q = heat required (in joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.184 J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
Since we know the volume of water (1.2 L) and the density of water (1 g/mL), we can determine the mass of water.
mass of water = volume of water * density of water
mass of water = 1.2 L * 1 g/mL = 1.2 kg = 1200 g
Now, we can calculate the heat required.
Q = (1200 g) * (4.184 J/g°C) * (100.0°C - 25.0°C)
Q = 503,040 J
Step 2: Determine the heat emitted by the LP gas combustion.
Given that 18% of the heat emitted by the LP gas combustion goes to heat the water, we can find the total heat emitted by the LP gas combustion.
Total heat emitted = heat required / (heat % going to water / 100)
Total heat emitted = 503,040 J / (18 / 100)
Total heat emitted = 2,794,667 J
Step 3: Convert the total heat emitted to kilojoules.
To match the given reaction's enthalpy change in kilojoules, we need to convert the total heat emitted to kilojoules.
Total heat emitted (kJ) = Total heat emitted (J) / 1000
Total heat emitted (kJ) = 2,794,667 J / 1000
Total heat emitted (kJ) = 2794.67 kJ
Step 4: Calculate the mass of LP gas required.
Using the equation ∆H = (mass of LP gas) × (ΔHrxn°) / (moles of LP gas), we can find the mass of LP gas required.
First, we need to calculate the number of moles of LP gas using the balanced equation's stoichiometry:
1 mole of C3H8(g) reacts to produce 3 moles of CO2(g)
Molar mass of C3H8 = 3 * (12.01 g/mol of C) + 8 * (1.01 g/mol of H) = 44.11 g/mol
moles of LP gas = mass of LP gas / molar mass of LP gas
mass of LP gas = ΔH / (moles of LP gas) × (45.11 g/mol)
mass of LP gas = (2794.67 kJ) / (-2044 kJ/mol) × (44.11 g/mol)
mass of LP gas ≈ 5.68 g
Therefore, approximately 5.68 grams of LP gas is required to heat 1.2 liters of water from 25.0°C to 100.0°C.