A student burned three different alcohols in turn to raise the temperature of 100 cm3 of water by 20 °C. Some of the results are shown in Table 4.1. The lar...

This question tests combustion of alcohols and how to read an energy comparison from a table. In every trial the burner delivered enough heat to raise the same 100 cm³ of water by the same 20 °C, so the useful heat gained by the water is the same each time, \( Q = m c \Delta T = 100 \times 4.18 \times 20 \approx 8360 \text{ J} \). What differs is the mass of alcohol needed to supply that heat, so the mass burned is the key to comparing the fuels.

(a) The flame for propan-1-ol is yellow / orange (a sootier, more luminous flame) [1]. Larger alcohol molecules have a higher proportion of carbon, so less oxygen reaches every carbon atom. Combustion is incomplete and tiny glowing carbon (soot) particles make the flame luminous yellow instead of clean blue.

(b) Propan-1-ol released the most energy per gram [1], because it needed the smallest mass (0.78 g) to produce the same 20 °C rise, so each gram must release the most energy [1]. Dividing the fixed heat by the mass burned makes this exact:

• methanol: \( 8360 \div 1.10 \approx 7600 \text{ J g}^{-1} \)
• ethanol: \( 8360 \div 0.90 \approx 9300 \text{ J g}^{-1} \)
• propan-1-ol: \( 8360 \div 0.78 \approx 10700 \text{ J g}^{-1} \)

A common error is to pick the largest mass burned; in fact the fuel that gives the same heating from the least fuel is the most energy dense.

(c) Any one sensible source of error, for example: different amounts of heat are lost to the surroundings for each alcohol; soot deposits on the can and stops heat transferring; the distance from the burner to the can is not kept the same; or draughts remove heat unevenly [1]. Each of these means the same 20 °C rise does not correspond to exactly the same energy released, so the comparison is not perfectly fair.