When backpacking in the wilderness, hikers often boil water to sterilize it for

Question

When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpacking trip and will need to boil 38L of water for your group.
What volume of fuel should you bring? Assume each of the following: the fuel has an average formula of C7H16 15% of the heat generated from combustion goes to heat the water (the rest is lost to the surroundings); the density of the fuel is 0.78 g/mL the initial temperature of the water is 25.0 degrees C and the standard enthalpy of formation of C7H16 is -224.4 kJ/mol. Answer in mL.

Explanation / Answer

Assuming you are boiling water at ground level (1.00 atm), where the boiling point is 100 C (the bp decreases with altitude as air pressure drops), the heat energy required to raise the temperature of 35L of water (density 1.00 kg/L) by 75 C is 35 kg x 4200 J/kgoC x 75oC = 11.025 MJ.

Only 15% of heat generated from combustion is useful, so
15% of the heat needed = 11.025 MJ
100% of heat needed = 100*11.025/15 = 73.5 MJ.

The heat released from burning the fuel in oxygen is calculated as follows :
7C + 8H2 ---> C7H16 releases 224.4 kJ/mol
therefore the reverse reaction C7H16 ---> 7C+8H2 requires 224.4 kJ/mol.
C + O2 ---> CO2 releases 393.5 kJ/mol
H2 + (1/2)O2 ---> H2O releases 285.8 kJ/mol

therefore the combustion of 1 mol of fuel
C7H16 + 7O2 + 4O2 ---> 7CO2 + 8H2O
releases 7*393.5 + 8*285.8 - 224.4 = 4816.5 kJ/mol = 4.82 MJ/mol of heat.

The amount of fuel required is 73.5 / 4.82 = 15.25 mol.

1 mol of fuel has a mass of 7*12 + 16*1 = 100 g so 15.25 mol has a mass of 1525 g.

The volume of fuel required is 1525g / 0.78 g/ml = 1955 ml = 1.955 L.

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