By the end of this lesson
- Carry out a safe food-calorimetry method.
- Calculate energy transferred to water and energy per gram.
- Evaluate uncertainty and systematic heat loss.
BIG QUESTION
Why does a burning food sample appear to contain less energy in a school experiment than its food label states?
Simple calorimetry measures energy reaching the water, not all chemical energy released, so systematic heat losses matter.
Measure mass and temperature change
Measure the food sample mass, place a known mass or volume of water in a boiling tube or metal calorimeter, and record its initial temperature. Ignite the food beneath the container and stir the water gently.
Record the highest water temperature and the remaining food mass if residue remains. The mass burned is initial minus final food mass.
Calculate energy transferred to water
Use energy transferred (J) = mass of water (g) × 4.2 J g⁻¹ °C⁻¹ × temperature change (°C).
Then calculate energy per gram = energy transferred ÷ mass of food burned. Divide by 1000 to convert J g⁻¹ to kJ g⁻¹.
Control the comparison
Keep water mass, starting temperature, distance from flame, container, stirring method and endpoint method constant. Use equal treatment rather than assuming equal-sized food pieces have equal mass.
Repeat each food type, identify anomalies using evidence and compare means. Replication reveals random variation but does not remove systematic heat loss.
Manage fire and burn hazards
Wear eye protection, tie back hair, use a heatproof mat and handle burning food with forceps. Keep flammable materials away and extinguish samples safely.
Do not eat laboratory food. Allow hot glassware and metal to cool before handling.
Interpret the estimate cautiously
Energy is lost to surrounding air, the container and clamp; combustion may be incomplete; the food may stop burning; and water may evaporate. The calculated value therefore usually underestimates total food energy.
A lid, draught shield, insulated calorimeter, smaller flame-to-container gap and oxygen-rich combustion can improve energy capture, though each change requires fresh risk assessment.
Variables and quality checks
A valid comparison needs both a calculation and a controlled energy-transfer setup.
| Element | Operational choice | Reason |
|---|---|---|
| Independent variable | Food type. | Compare different energy stores. |
| Dependent variable | Calculated J g⁻¹ or kJ g⁻¹. | Corrects for different masses burned. |
| Control variables | Water mass, gap, container and stirring. | Keep heat capture comparable. |
| Repeats | At least three usable trials per food. | Estimate variability and a mean. |
| Safety | Goggles, forceps, heatproof mat, tied hair. | Reduce burn, fire and eye risk. |
Language that earns precision
- calorimetry
- Estimation of energy transfer by measuring a temperature change.
- specific heat capacity
- Energy needed to raise 1 g of a substance by 1 °C.
- temperature change
- Final temperature minus initial temperature.
- systematic error
- A consistent bias shifting measurements in one direction.
- energy per gram
- Energy estimate divided by the mass of sample burned.
Common misconceptions
Claim: The largest temperature rise proves the food has the most energy per gram.
Correction: Account for water mass and mass of food burned.
Claim: Repeats eliminate heat loss.
Correction: Repeats reduce random uncertainty; systematic losses remain.
Claim: The water receives all energy released.
Correction: Much energy heats the apparatus and surroundings.
Claim: Food that will not relight has fully combusted.
Correction: Residue and incomplete combustion may remain.
Match the answer to the command word
Show ΔT first, use mass burned, then convert units only at the end.
Name where energy goes and state the direction of bias.
Match each change to a specific loss or uncertainty and include safety.
Recall, apply, then justify
A food sample heats 20.0 g of water from 20.0 °C to 47.0 °C. Calculate the energy transferred to the water.
Hint
Use m × 4.2 × ΔT.
Suggested marking points
- Temperature change = 27.0 °C.
- Energy = 20.0 × 4.2 × 27.0.
- Energy transferred = 2268 J, or 2.27 kJ to three significant figures.
In the same trial, 0.35 g of food burned. Calculate the energy transferred per gram in kJ g⁻¹.
Hint
Divide the previous answer by 0.35, then convert to kJ.
Suggested marking points
- 2268 J ÷ 0.35 g = 6480 J g⁻¹.
- Convert by dividing by 1000.
- Energy estimate = 6.48 kJ g⁻¹.
Explain why the calculated value is likely below the label value.
Hint
Identify two destinations or failures and the direction of effect.
Suggested marking points
- Energy is transferred to air/apparatus rather than only to water.
- Combustion may be incomplete.
- Water evaporation or unburned food may reduce the measured rise.
- Therefore the water-temperature calculation underestimates chemical energy released.
Design an improved comparison of three foods and explain how you would judge both validity and reliability.
Planning hint
Include variables, mass correction, repeats, safety and systematic error.
What a strong design includes
- Measure initial and final food mass to obtain mass burned.
- Use a fixed water mass, container and flame distance.
- Stir and measure temperature consistently.
- Use insulation, a lid or draught shield while maintaining oxygen supply.
- Repeat each food and compare means plus spread.
- Calculate kJ g⁻¹ rather than raw temperature rise.
- Use eye protection, forceps, heatproof mat and controlled extinguishing.
- State that improved precision does not guarantee removal of heat-loss bias.