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Choosing a Fuel for a Water-Heating Demonstration: Energy, Cost and Practical Judgement

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Choosing a Fuel for a Water-Heating Demonstration: Energy, Cost and Practical Judgement

A school team must heat 500 g of water from 28 °C to 90 °C for a demonstration and is allowed to buy only one fuel. Three options are available locally:

  • Ethanol: ΔHc = −1376 kJ mol−1, Mr = 46, RM 45 per kg, sold in a plastic bottle.
  • Methanol: ΔHc = −726 kJ mol−1, Mr = 32, RM 18 per kg, also in a bottle.
  • Butane: ΔHc = −2877 kJ mol−1, Mr = 58, RM 30 per kg, sold in a sealed pressurised cartridge for a small portable stove.

The team's spirit burner transfers only 40% of the heat released to the water; the remainder is lost to the surroundings and the pot. The teacher asks the team to justify the purchase with calculations rather than impressions, and to state one assumption in those calculations that could fail in practice. [Specific heat capacity of water = 4.2 J g−1 °C−1]

1. Calculate the energy required and the mass of each fuel needed for the demonstration.

Read the worked answer

Energy actually taken up by the water

Q = mcΔθ = 500 × 4.2 × (90 − 28) = 500 × 4.2 × 62 = 130 200 J = 130.2 kJ

Energy that must be released by the fuel

Only 40% reaches the water, so release = 130.2 ÷ 0.40 = 325.5 kJ

Moles and mass of each fuel

Ethanol: 325.5 ÷ 1376 = 0.2366 mol → 0.2366 × 46 = 10.9 g

Methanol: 325.5 ÷ 726 = 0.4483 mol → 0.4483 × 32 = 14.3 g

Butane: 325.5 ÷ 2877 = 0.1131 mol → 0.1131 × 58 = 6.56 g

Note the pattern: butane releases far more energy per mole, so the smallest mass is needed. Ethanol's mole releases nearly twice the energy of a mole of methanol, and also carries less mass per mole than butane, which is why it lands in between.

2. Calculate the cost of each fuel for the demonstration and identify the cheapest option.

Read the worked answer

Converting grams to kilograms and costing

Ethanol: 10.9 g = 0.0109 kg → 0.0109 × RM 45 = RM 0.49

Methanol: 14.3 g = 0.0143 kg → 0.0143 × RM 18 = RM 0.26

Butane: 6.56 g = 0.00656 kg → 0.00656 × RM 30 = RM 0.20

Ranking

Cheapest is butane (RM 0.20), then methanol (RM 0.26), then ethanol (RM 0.49). The gap between butane and ethanol is about RM 0.29 for a single demonstration.

Check the reasoning

Cost per mole of fuel is not the useful comparison; cost per mole of fuel divided by energy released per mole is, because what the team is buying is heat. Ethanol is the most expensive here despite being a 'clean' fuel, because its cost per kilogram is the highest and it is the least energy-dense of the three per gram. For a single demonstration the money difference is small, so the decision should not rest on cost alone.

3. Recommend one fuel, taking into account factors beyond cost, and state one assumption in the calculations that could fail in practice.

Read the worked answer

Factors beyond cost

  • Apparatus: butane needs a pressurised cartridge and a stove; the stove is a capital cost not included in RM 30 per kg, so the true cost is higher than calculated. Ethanol and methanol run in an ordinary spirit burner.
  • Safety: methanol is toxic by skin absorption and inhalation and can cause blindness; it also burns with a nearly invisible flame, so a spill fire is easy to miss. Ethanol is the safest of the three to handle and store in a school. Butane cartridges are pressurised and must not be heated.
  • Reliability: butane is the lightest fuel to carry and gives a controllable, hot flame; ethanol gives a visible flame that is good for a demonstration.

Recommendation

For a school demonstration, choose ethanol: it is the safest to handle, needs no special apparatus, gives a visible flame, and the extra RM 0.29 is trivial. If the demonstration were repeated many times and a stove were already available, butane would be the rational choice on cost and mass.

Assumption that could fail

The calculation assumes complete combustion and a constant 40% heat transfer. In practice, a yellow smoky flame means incomplete combustion producing soot and carbon monoxide, which releases far less heat than the quoted ΔHc, and heat loss rises as the flame plays on the sides of the pot, so more fuel than calculated may be needed.