Whenever organic matter is decomposed under oxygen free conditions, methane is one of the products. Thus enormous deposits of natural gas, which is almost entirely methane, serve as a major source of fuel for home and industry (dHrxn for the combustion of CH4 = -802 kJ/mol.) What volume (in ft^3) of natural gas, measured at STP, is required to heat 1.00 qt of water from 25.0 degrees C to 83.50 degrees C (d of H2O = 1.00 g/mL; d of CH4 at STP = 0.72 g/L)?
I got 0.228 cubic feet.....is this right?
I obtained 0.226; I suspect we just didn't use the same conversion fators for one of the conversions.
Thank you so much for all your help!
To solve this problem, we need to calculate the heat energy required to heat the water and then find the volume of natural gas needed to produce that amount of energy.
Step 1: Calculate the mass of water.
Given: Density of water (d H2O) = 1.00 g/mL
Volume of water = 1.00 qt = 946 mL
Mass of water = Volume × Density = 946 mL × 1.00 g/mL = 946 g
Step 2: Calculate the heat energy required to heat the water.
Given: Change in temperature (ΔT) = 83.50 - 25.0 = 58.5 °C
Specific heat capacity of water (c) = 4.18 J/g°C
Heat energy (Q) = Mass × ΔT × c
Q = 946 g × 58.5 °C × 4.18 J/g°C = 226,183 J
Step 3: Convert the heat energy to kilojoules.
1 kJ = 1000 J
Q = 226,183 J ÷ 1000 = 226.183 kJ
Step 4: Use the thermochemical equation to determine the volume of natural gas.
Given: ΔHrxn for the combustion of CH4 = -802 kJ/mol
Density of CH4 at STP (d CH4) = 0.72 g/L
1 mole of CH4 releases 802 kJ of heat energy.
Molar mass of CH4 = 12.01 g/mol (C) + 4(1.01 g/mol) (4H) = 16.04 g/mol
The number of moles of CH4 = Mass ÷ Molar mass
moles of CH4 = 226.183 kJ ÷ 802 kJ/mol = 0.2815 mol
Step 5: Calculate the volume of CH4 at STP.
Given: Density of CH4 at STP = 0.72 g/L
Convert moles of CH4 to grams:
Mass of CH4 = moles × Molar mass = 0.2815 mol × 16.04 g/mol = 4.51 g
Convert grams to liters:
Volume = Mass ÷ Density = 4.51 g ÷ (0.72 g/L) = 6.26 L
Step 6: Convert liters to cubic feet.
1 L = 0.0353 ft^3
Volume in cubic feet = 6.26 L × 0.0353 ft^3/L = 0.221 ft^3
Therefore, the correct volume of natural gas required to heat 1.00 qt of water is 0.221 cubic feet (rounded to three decimal places).
To determine the volume of natural gas required to heat water, we need to calculate the energy needed to heat the water and then convert that energy into the amount of natural gas required.
First, let's calculate the heat energy required to heat the water using the formula:
q = m * c * ΔT
Where:
q = heat energy
m = mass of water
c = specific heat capacity of water
ΔT = temperature change
Given:
Mass of water (m) = 1.00 quart = 946 grams (since 1 qt = 946 mL and the density of water is 1.00 g/mL)
Specific heat capacity of water (c) = 4.18 J/g°C (assuming units are in J/g°C)
Temperature change (ΔT) = 83.50°C - 25.0°C = 58.50°C
q = 946 g * 4.18 J/g°C * 58.50°C
q = 237,447.12 J
Next, let's convert the heat energy into kJ using the conversion factor:
1 J = 0.001 kJ
Q = 237,447.12 J * 0.001 kJ/J
Q = 237.45 kJ
Now, let's calculate the moles of CH4 (methane) required using the equation:
dH = n * dHrxn
Where:
dH = heat energy (in this case, 237.45 kJ)
n = moles of CH4
dHrxn = enthalpy change for the combustion of CH4 (-802 kJ/mol)
237.45 kJ = n * (-802 kJ/mol)
n = 237.45 kJ / (-802 kJ/mol)
n ≈ -0.296 mol
Since mole quantities can't be negative, we take the absolute value of n:
n = 0.296 mol
Now, let's convert the moles of CH4 into volume using the ideal gas law equation:
PV = nRT
Where:
P = pressure (STP conditions = 1 atm)
V = volume of gas (what we want to find)
n = moles of gas (0.296 mol)
R = gas constant (0.0821 atm·L/mol·K)
T = temperature (STP conditions = 273.15 K)
Plugging in the values:
(1 atm) * V = (0.296 mol) * (0.0821 atm·L/mol·K) * (273.15 K)
V = (0.296 mol) * (0.0821 atm·L/mol·K) * (273.15 K) / 1 atm
V ≈ 6.73 L
Next, let's convert the volume from liters (L) to cubic feet (ft^3):
1 L = 0.0353 ft^3
V = 6.73 L * 0.0353 ft^3/L
V ≈ 0.238 ft^3
Therefore, the volume of natural gas required to heat 1.00 quart of water is approximately 0.238 cubic feet (ft^3), which is close to the value you obtained of 0.228 cubic feet (ft^3).