Fig. 4.1 shows the apparatus set out on your bench for burning a liquid fuel to heat water. The two metal plates labelled C stand one on each side of the fl...

This is a combustion calorimetry experiment: a burning fuel heats a known mass of water, and from the temperature rise you find the energy released. The reasoning throughout is the energy equation \( E = mc\,\Delta T \) and the idea of a fair, low-loss measurement.

(a) [2] A is the spirit (alcohol) burner [1] that holds and burns the liquid fuel, and B is the clamp and stand (boss and clamp) [1] that holds the can steady above the flame.

(b) [2] To pass as much heat as possible into the can, clamp the can so its base is just above the tip of the flame, about 4 cm to 5 cm above the wick [1], and stand it directly over the wick with the two plates close on each side so draughts cannot blow the flame away from the base [1]. Being close to the flame and shielded from moving air both cut the heat lost to the surroundings.

(c) [2] The volume of water is measured with a measuring cylinder (a 250 cm³ one suits 200 cm³) [1]. The starting temperature must be recorded to the nearest 0.5 ℃ with its unit, e.g. 19.5 ℃ [1], matching the precision the thermometer allows.

(d) [1] Before lighting, record the mass of the burner and its cap to 0.01 g, the precision of a laboratory balance; this is the "before" mass for the fuel used.

(e) [2] When the water has risen by exactly 25.0 ℃, put out the flame and record the burner and cap mass again to 0.01 g; it is lower than the first mass [1]. The mass must be taken straight away because the volatile fuel keeps evaporating from the wick, so any delay loses extra mass and makes the mass of fuel burned look too large [1].

(f) [2] Use \( E = mc\,\Delta T \) with \( m = 200 \text{ g} \) (since \( 1 \text{ cm}^3 \) of water has mass \( 1 \text{ g} \)), \( c = 4.2 \text{ J g}^{-1}\,{}^{\circ}\text{C}^{-1} \) and \( \Delta T = 25.0\,{}^{\circ}\text{C} \): \[ E = 200 \times 4.2 \times 25.0 \; [1] = 21\,000 \text{ J} = 21 \text{ kJ} \; [1] \] Correct substitution earns the first mark and the converted answer in kJ the second.

(g) [2] Energy per gram of fuel is the energy transferred divided by the mass of fuel burned, \( \dfrac{21\,000 \text{ J}}{\text{mass burned}} \) [1], quoted with a unit. For example, if 1.05 g burned, \( \dfrac{21\,000}{1.05} = 20\,000 \text{ J/g} = 20.0 \text{ kJ/g} \) [1] (your own mass gives your own value).

(h) [2] With the plates C removed, more fuel has to burn to give the same 25.0 ℃ rise [1], because moving air now carries the hot gases away from the can, so more heat is lost to the surroundings [1]. This is why shielding the flame improved the result in part (b).

(i) [1] Give any one sensible safety precaution while the burner is alight, for example tie back long hair and loose clothing, wear eye protection, never refill the burner while it is lit, or keep the stock bottle of fuel well away from the flame.

Exam tip: always convert to the unit the question asks for. Here the energy calculation naturally comes out in joules, so remember the final "in kJ" step by dividing by 1000.