Answer:
$0.0224
Explanation:
First, determine the amount of heat required to raise the temperature of 450 mL water from 20°C to 100°C.
density of water = 1 g/mL
Use = AH = m(AT)Cp
AH (delta H) = amount of heat required
m = mass of water = 450 mL: (1g/mL) = 450 g
AT(delta T) = temperature difference = 100 - 20 = 80 °C
Cp = specific heat of water (from a table) = 4.18 J/g °C
So,
AH = 450 (80) (4.18) = 150 480 J = 150.48 kJ
Second, determine the heat ACTUALLY required
Since 50 % of heat produced (from combustion of methane) are used for water,
heat ACTUALLY required = 2 x 150.48 kJ = 300.96 kJ
Third, determine the amount of methane in the supply (in mole)
Assume ideal gas,
PV = nRT
P = 1.0 atm
V = 17 ft3 = 481.38 L
T = 20 + 273 = 293 K
R = 0.082 1 L atm K−1 mol−1
n = PV/RT
so,
n = (1 x 481.38)/(293 x 0.0821) = 20.011 mol
Interpretation: 17 ft3 of methane supplied is equivalent to 20.011 mol methane (17 ft3/20.011 mol)
Fourth, determine the heat produced from combustion of methane
The combustion of methane gas is represented by the reaction:
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
heat of combustion =
[heat of formation of CO2 (g) + 2 x heat of formation of H2O (g)] -
[heat of formation of CH4 (g) + 2 x heat of formation of O2 (g)]
From the standard table:
= [ -393.5 kJ + 2x(-241.8 kJ)] - [-74.8 kJ + 2x(0 kJ)]
= - 802.3 kJ
Interpretation: The combustion of 1 mole of methane produces 802.3 kJ, otherwise known as molar enthalpy (-802.3 kJ/mol)
Fifth, determine the amount of methane required from energy requirement for heating water
heat required = 300.96 kJ
So,
methane required = 300.96 kJ (1mol/ 802.3 kJ) = 0.3751 mol
Sixth, determine the volume of methane required
Volume methane required = 0.3751 mol (17 ft3/20.011 mol)
= 0.3186 ft3
Finally, the cost required at price rate $1.20/17 ft3
Cost of methane = 0.3186 ft3 ($1.20/17 ft3) =
$0.0224
