sigfigs Conversion

Question

A backpacker wants to carry enough fuel to heat 2.8  of water from 27  to 100.0 .

If the fuel he carries produces 36

of heat per gram when it burns, how much fuel should he carry? (For the sake of simplicity, assume that the transfer of heat is 100 

 efficient.)

Answer 1

The units were stripped out by the posting software, but they can be recovered - and showing that they can is worth doing, because it is a real skill.

Recovering the units

2.8 of water. Grams is out: 2.8 g is barely a teaspoon, and nobody carries fuel for that. Kilograms and litres both work, and for water they give the same number since 1 L has a mass of 1 kg. So take 2.8 kg = 2800 g.

27 and 100.0. These have to be degrees Celsius. 100.0 is the boiling point of water, which is what a backpacker is heating toward, and 27 degrees C is a plausible ambient temperature.

36 of heat per gram. This one settles itself at the end. If it were 36 J/g you would need about 24 kg of fuel, which nobody is carrying up a mountain. At 36 kJ/g the answer is a couple of dozen grams, which is right - and 36 kJ/g is the correct order of magnitude for a hydrocarbon camping fuel.

So: heat 2800 g of water from 27 degrees C to 100.0 degrees C, with fuel giving 36 kJ per gram.

Step 1 - heat the water needs

delta T = 100.0 - 27 = 73 degrees C

q = m c delta T = (2800 g)(4.184 J/g degree C)(73 degrees C)

q = 8.55 x 10^5 J = 855 kJ

Step 2 - convert heat to fuel mass

The energy density is a conversion factor. Write it so that kJ cancels:

855 kJ x (1 g fuel / 36 kJ) = 23.8 g

He should carry about 24 g of fuel.

Significant figures

Since your title flags sig figs, here is the reasoning. 2.8 has two, 36 has two, and 100.0 has four. The subtraction 100.0 - 27 is governed by decimal places rather than significant figures, and 27 has none past the decimal point, so delta T = 73 with two significant figures. Two is therefore the limit throughout, and 24 g is the correct report - not 23.8 g and certainly not 23.76 g.

A check worth making

Note the problem says to assume 100 percent heat transfer. A real camp stove delivers maybe 30 to 50 percent of its heat to the pot, the rest going into the surrounding air, so a real backpacker would want roughly 50 to 80 g. If a later part of the problem gives you an efficiency, remember to divide by it - lower efficiency means more fuel, never less.

Also note this only brings the water to 100 degrees C; it does not boil any of it away. Vaporizing even a tenth of that water would take another 630 kJ, comparable to everything calculated above, which is why you take a pot off the heat as soon as it boils.