Calculate the volume of natural gas (methane) required to bring 2.6 L of water to a boil. Assume the initial temperature of the water is 21 C, and the methane is at 1 bar and 25 C. Assume that methane is an ideal gas, the specific heat capacity of water to be constant and equal to 4.184 J/g/K over this temperature range, and the density of water to be 1.0 g/cm3. The combustion of methane is given by
CH4 + 2 O2 �¨ CO2 + 2 H2O ƒ¢Hcomb = -890.36 kJ/mol
I responded to a similar problem, different numbers I think but the same concept, a day or so ago and here is my response.
http://www.jiskha.com/display.cgi?id=1322859803
To calculate the volume of natural gas (methane) required to bring 2.6 L of water to a boil, we need to follow several steps.
Step 1: Calculate the amount of heat required to raise the temperature of the water from 21 °C to its boiling point.
The specific heat capacity of water is given as 4.184 J/g/K, and the density of water is 1.0 g/cm³. Since we have 2.6 L of water, we need to calculate the mass of water using its density:
Mass of water = Density × Volume of water
= 1.0 g/cm³ × 2.6 L
= 2600 g
The temperature change is from 21 °C to the boiling point of water, which is 100 °C. The heat required can be calculated using the formula:
Heat required = mass × specific heat capacity × temperature change
= 2600 g × 4.184 J/g/K × (100 °C - 21 °C)
Step 2: Convert the heat required to the energy unit used in the given enthalpy of combustion.
The enthalpy of combustion of methane is given as -890.36 kJ/mol. To convert the heat required to kJ, we need to divide by Avogadro's number (to convert grams to moles) and multiply by the molar enthalpy of combustion:
Heat required = (2600 g / molar mass of water) × (1 mol water / 18.015 g) × (-890.36 kJ/mol)
Step 3: Convert the heat required to the volume of methane gas.
The enthalpy of combustion of methane is -890.36 kJ/mol, and it produces 2 moles of water. Therefore, the molar enthalpy of combustion for water is -890.36 kJ/mol / 2 = -445.18 kJ/mol.
To find the volume of methane gas, we need to divide the heat required by the molar enthalpy of combustion for water:
Volume of methane gas = (Heat required / molar enthalpy of combustion for water) × molar volume of gas at given conditions
The molar volume of a gas at standard temperature and pressure (STP) is approximately 22.4 L/mol. However, our methane gas is at 1 bar and 25 °C, so we need to correct the molar volume using the ideal gas law:
P × V = n × R × T
Where:
P is the pressure (1 bar = 100,000 Pa)
V is the volume (molar volume at STP = 22.4 L/mol)
n is the number of moles (unknown)
R is the gas constant (8.314 J/(mol·K))
T is the temperature (25 °C = 298 K)
We can rearrange the equation to solve for n:
n = (P × V) / (R × T)
Finally, we can substitute this value of n into the previous equation to find the volume of methane gas.
Note: Make sure to convert the final result to the desired unit of volume.
Now you can use the above steps to calculate the volume of methane gas required to bring 2.6 L of water to a boil.