Mixed tasks often start with a biology idea, such as the energy stored in food, and then ask for a physics measurement, such as the temperature rise of water. Your job is to trace the energy: store, transfer, measurement, loss.
This lesson is part of biological mechanisms in mixed tasks. It discusses a food energy investigation conceptually only. Any real version of this practical is done under teacher supervision following your school’s safety rules.
What is the energy chain?
Food contains stored chemical energy. When a food sample is burned, that energy is transferred as heat to the surroundings, and some warms a known mass of water.
The energy gained by the water is found with the equation energy (J) = mass of water (g) × specific heat capacity (J/g°C) × temperature rise (°C). Then, energy per gram of food = energy transferred ÷ mass of food burned.
Worked example (invented data)
In a teaching example, a food sample of mass 0.50 g heats 50 g of water. The water temperature rises by 20 °C. Use 4.2 J/g°C for water.
Step 1, energy gained by the water: 50 × 4.2 × 20. First 50 × 4.2 = 210. Then 210 × 20 = 4200 J.
Step 2, energy per gram of food: 4200 ÷ 0.50 = 8400 J/g, which is 8.4 kJ/g.
Step 3, check the unit conversion: 8400 ÷ 1000 = 8.4 kJ/g. Check by multiplying back: 8.4 × 0.50 = 4.2 kJ = 4200 J, which matches Step 1.
Step 4, evaluate: this teaching value is lower than the energy a data book gives for most foods. The reason is that much of the heat warms the air and the equipment, not the water, and the sample may burn incompletely. The result is an underestimate, not a sign that the food holds less energy.
The mistake to watch for
Mistaken answer: energy = 0.50 × 4.2 × 20 = 42 J
The student used the mass of the food instead of the mass of the water. The heat capacity equation uses the mass of the substance that is being heated. The food mass is used only in the last step, to find energy per gram.
A second slip is mixing J and kJ. Write the unit on every line.
Check yourself
1. 100 g of water rises by 10 °C. How much energy did the water gain? (Use 4.2 J/g°C.)
Show answer
100 × 4.2 = 420, then 420 × 10 = 4200 J.
2. A 0.30 g sample heats 30 g of water by 12 °C. Find the energy per gram of food.
Show answer
Energy to water = 30 × 4.2 × 12 = 126 × 12 = 1512 J. Energy per gram = 1512 ÷ 0.30 = 5040 J/g. Check: 5040 × 0.30 = 1512.
3. Give one reason the calculated value is lower than the true energy in the food.
Show answer
Heat is lost to the surrounding air and the equipment, so not all the energy reaches the water. Incomplete burning is another valid reason.
Where this leads next
Move on to using a food-web change to test an evidence-based conclusion. The scientific investigation critic is useful for deciding which improvements to the energy method are sensible, and the mixed practice set has more calculations.
If calculations in a biology setting are where you hesitate, our teachers can support you in online one-to-one Combined Science tuition.