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

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 33 L 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 C and the standard enthalpy of formation of C7H16 is -224.4 kJ/mol.

Answer 1

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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 33L of water (density 1.00 kg/L) by 75 C is 33kg x 4200J/kgC x 75C = 10.4 MJ.

Only 15% of heat generated from combustion is useful, so

15% of the heat needed = 10.4 MJ

100% of heat needed = 100*10.4/15 = 69.33MJ.

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 = 4.82 MJ of heat.

The amount of fuel required is 69.33 / 4.82 = 14.4 mol.

1 mol of fuel has a mass of 7*12+16*1=100g so 14.4 mol has a mass of 1,440 g.

The volume of fuel required is 1440g / 0.78 g/ml = 1846 ml = 1.85L.

Source(s): http://chemed.chem.wisc.edu/chempaths/GenChem-Text...