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 40L 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

Explanation / Answer

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 In such an involved problem, I might have made an error, as this to me seems like a lot of fuel. Recheck my math to be sure.

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