A student heats 0.30 kg of olive oil in a metal pan using a gas burner. The initial temperature is 22 °C and the desired cooking temperature is 180 °C. The ...

(a) Thermal energy required to heat the oil [2]

Using the thermal energy equation with the specific heat capacity of olive oil:

\( E = mc\Delta\theta = 0.30 \times 1970 \times (180 - 22) \) [1]

\( E = 0.30 \times 1970 \times 158 = \mathbf{93\,378 \text{ J}} \) (accept 93 400 J or 93.4 kJ) [1]

Note that the specific heat capacity of oil (1970 J/(kg°C)) is less than half that of water (4200 J/(kg°C)), meaning oil requires less energy per kilogram per degree to heat up. However, the very large temperature rise (158°C) still requires a substantial amount of energy.

(b) Minimum time to reach 180°C [2]

The gas burner delivers energy at an effective rate of 800 W. The minimum time assumes all this energy goes into heating the oil:

\( t = \frac{E}{P} = \frac{93\,378}{800} \) [1]

\( t = \mathbf{116.7 \text{ s}} \) (accept 117 s or approximately 1.9 minutes) [1]

(c) Why the actual time is longer [1]

Some energy is lost to the surroundings from the pan by convection (hot air rising from the pan) and radiation (infrared emission from the hot surfaces), so the effective heating rate reaching the oil is less than 800 W. [1] The pan itself also absorbs some energy as its temperature rises.

(d) Why oil is used instead of water for frying above 100°C [2]

Cooking oil has a much higher boiling point than water (well above 180°C for olive oil), so it remains in the liquid state at typical frying temperatures. [1] Water, on the other hand, boils at 100°C at atmospheric pressure and would simply evaporate away long before reaching 180°C, making it impossible to use as a frying medium at high temperatures. [1] The high boiling point of oil allows it to transfer heat efficiently to food at the elevated temperatures needed for frying.