2.5.5Chemical Bonding, Application Of Chemical Reactions and Organic Chemistry
Energy Density of a Fuel
5 min read 7 parts Lesson 45 of 54 Unit quiz
Section 1 of 7Why measure the energy from a fuel
Comparing fuels
- Comparing fuels Different fuels release different amounts of energy for the same mass burned, so we need a fair experiment to compare them.
- The energy released per gram of fuel burned is a useful way of ranking fuels.
Calorimetry
- Calorimetry Calorimetry is the technique of using the temperature rise of a known mass of water to work out how much energy was transferred to it.
- The fuel burns in a spirit burner under a copper can of water, because copper conducts heat well.
Section 2 of 7Method
1
Measure a fixed volume of water, for example 200 cm3, into a copper can (calorimeter) and clamp it above the burner.
2
Weigh the spirit burner containing the fuel on a balance and record its mass.
3
Measure the starting temperature of the water with a thermometer.
4
Light the burner and stir the water gently all the time so the temperature is even throughout.
5
When the temperature has risen by about 20 °C, put out the flame, using the burner's cap.
6
Record the highest temperature reached and then re-weigh the burner. The decrease in mass is the mass of fuel burned.
Section 3 of 7Controlling the experiment
Fair test
Variables
- Variables The independent variable is the type of fuel.
- The dependent variable is the temperature rise.
- Control the volume of water, the starting temperature, the can and the distance between flame and can.
Heat loss
- Heat loss Much of the energy escapes into the air, so a lid on the can and a draught shield reduce losses.
- The calculated value is always lower than the true energy released.
Incomplete combustion
- Incomplete combustion A sooty can means some fuel did not burn completely, so less energy was released than possible.
Section 4 of 7Calculating the heat energy change
| Quantity | Symbol | Unit |
|---|---|---|
| Heat energy change | Q | joules, J |
| Mass of substance heated (the water) | m | grams, g |
| Specific heat capacity | c | J/g/°C (for water 4.18) |
| Temperature change | ΔT | °C |
Section 5 of 7Worked example
1Write the equation
- Write the equation Q = m × c × ΔT. The equation is provided in the exam, but you must be able to use it.
2Substitute
- Substitute A burner containing 1.023 g of propanol heats 200 g of water by 30 °C, so Q = 200 × 4.18 × 30.
3Calculate
- Calculate Q = 25 080 J of energy transferred to the water.
4Interpret
- Interpret The fuel released this energy, so its energy change is negative (−25 080 J), showing an exothermic reaction.
- Dividing by the mass burned gives about 24.5 kJ per gram.
Section 6 of 7Key terms
| Term | Meaning |
|---|---|
| Calorimeter | A container, usually a copper can of water, used to measure energy transferred in a reaction. |
| Specific heat capacity | The energy needed to raise the temperature of 1 g of a substance by 1 °C; for water it is 4.18 J/g/°C. |
| Energy density | The amount of energy released by a fuel per unit mass (or volume) burned. |
| Exothermic | A reaction that releases energy to the surroundings, giving a negative energy change. |
| ΔT | The change in temperature, final temperature minus starting temperature. |
Section 7 of 7Exam technique
1
Using the mass of the fuel as m in Q = mcΔT. The mass is that of the water being heated.
2
Forgetting that heat is lost to the surroundings, so the value from the experiment is lower than the real energy released.
3
Not stirring the water, which gives an uneven temperature and an unreliable ΔT.
4
Omitting the sign or unit. Write the answer in joules and show that an exothermic change is negative.