When backpacking in the wilderness hikers often boil water

- 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.<?xml:namespace prefix = o ns = \"urn:schemas-microsoft-com:office:office\" />

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 degre celcius and the standard enthalpy of formation of C7H16:;is -224.4Kj/mol.

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Solution

First find amount of heat energy needed.

Mass water in grams = 30000. (one liter water weighs 1000 grams.

Temp increase from 25 to l00 degrees C (boiling point)
= 75 degrees C

Heat energy needed = 75 degrees x 30000 grams x lcal/gm water specific heat =
a total of 2.25 x 10 to 6 th power calories

Since only l5% of the heat generated goes to heat the water, then to find total heat energy needed (X)
we say that 2.25 x l0 to the 6th power = .l5X

so X = 2.25 x l0 to sixth divided by .15 to get total of
1.5 x 10 to 7th power calories. or l.5 x l0 to 4th kcal.

Now to get moles of fuel needed just divide all this heat energy by the heat delivered per mole of fuel burned, given as 224.4 kj/mol

Since; we have calculated the total heat energy needed in Kcal we need to get our enthalpy of combustion of one mole of heptane (C7Hl6) into kcal.

Using conversion table: l kcal = 4.184 kj

So 1.5 x 10 to the 4th kcal x 4.184 kl/kcal =6.28 x 10 to the 4th kj total heat energy needed.

Now to get moles of fuel just dicide total heat energy by heat output per mole fuel (224.4kj)

So 6.28 x 10 to fourth divided by 224.4 =280 moles fuel

Now add up the weight of one mol heptane
7x12 +16x1.008= 100.13 g/mol fuel

100.13g/mol x 280 moles fuel =2.8 x 10 to 4th grams fuel

now divide mass fuel by its density to get volume fuel

so 2.8 x 10 to 4th gms divided by .78g/ml =3.59 x 10 to 4th ml or 35.9 liters fuel