UK GCSE level age ~14-16, ~US grades 9-10 Biology revision notes re-edit 22/05/2023 [SEARCH]

Respiration: 8. Simple experiments to measure the release of energy on burning foods like fats

Doc Brown's Biology exam study revision notes

*

There are various sections to work through,

after 1 they can be read and studied in any order.

*

(8) Simple experiment to measure the release of energy on burning foods like fats

Foods such as animal fats, vegetable oils and carbohydrates like glucose and starch are concentrated chemical energy stores.

They are metabolised in the body to power all of a cell's chemistry and provide thermal energy to warm blooded creatures like ourselves!

You do a controlled burning of a lump of food to get an idea of how much chemical energy it contains by converting .

A bit of burning food calorimetry (experimental set-up illustrated on the right).

A very simple investigation method to measure the calorific value of food

Wear eye protection and take care near flames.

Add exactly 20 cm3 of water to a pyrex boiling tube (better than narrow test tube).

The boiling tube is clamped above the lab bench in an inclined position.

A 0-100oC thermometer is carefully placed in the water.

You can use any dried food e.g. beans, bread, nuts or pasta and a weighed lump of it is skewered onto the end of a mounted needle.

The water is gently stirred with the thermometer and the initial temperature recorded.

The food lump is ignited with a bunsen burner flame and then held with a steady had under the bottom of the boiling tube.

If the flame goes out, relight the food, repeating this until it has all burned away (or the residue no longer burns).

When all the food has burned away, re-stir the water gently with the thermometer and note the final higher temperature.

Calculation of energy content

The heat released = mass of water (g) x temperature change (oC) x specific heat capacity of water.

e.g. 0.75 g nut was burned as much as possible.

The initial and final temperatures were 21oC and 43oC. Temperature rise = 43 - 21 = 22oC.

20 cm3 of water is 20 g (density = 1.0 g/cm3).

Specific heat capacity of water is 4.18 J/kgoC. (see GCSE physics notes for more on )

Therefore heat released = 20 x 22 x 4.18 = 1839 J

The result can then be expressed in terms of energy content per unit mass of food e.g. J/g.

2500 J/g (2 sf) or 2.5 kJ/g or 2500 kJ/kg

You can think of this calorific value as a measure of energy density.

Experimental errors

This experiment is NOT very accurate at all, but it gives a rough estimate of the energy content of a food.

(i) Heat is being continuously lost from the boiling tube which isn't insulated.

(ii) Even more heat is lost by the warm convecting flame gases rising beyond the boiling tube, because you cannot collect all the thermal energy from the hot flame gases.

(iii) You cannot burn every bit of food, there is always a burnt out carbonised residue.

(iv) The yellowish flame tends to be sooty, so all the carbon is not oxidised to carbon dioxide.

For commercial calorific values scientists use a sealed and well insulated calorimeter, with no heat lost, you get very accurate values of energy content. The instrument is called a bomb calorimeter and pure oxygen is used - kapow!.

Further experiments

Repeat with different foods, but keep the volume of cool start water the same.

Repeat the calculation and compare the calorific values and energy densities of different foods.

You can also look up the molecular structure of some of the molecules in the food e.g.

the above diagram shows a section of a vegetable oil molecule (a long chain fatty acid), and as you can see, there are lots of carbon and hydrogen atoms to be oxidised to carbon dioxide and water respectively.

Fat molecules have a higher energy density than carbohydrates because the latter contains a greater proportion of oxygen atoms - so a greater proportion of the carbon atoms are partially oxidised, so less energy can be released on burning or by metabolic chemistry in the body.

Using a more accurate calorimeter (see energy changes in chemistry)

You can use a more sophisticated calorimeter system (illustrated on the right).

You can use this system employing a copper can (good heat conductor) to hold the water and employ a draught shield to minimise heat losses by convection..

You fill the little wick burner with vegetable oil and weigh it.

Pour in 100 g (~100 cm3) of water into the copper calorimeter and measure its temperature.

Light the burner and place it carefully under the suspended calorimeter (clamp not shown).

Burn for 5 minutes, extinguish the flame and take the final increased temperature of the water after stirring it to get the average temperature of the bulk liquid.

Suppose the burner containing the vegetable oil weighed 20.55 g.

After burning it weighed 20.05 g and the temperature rose from 24.5oC to 40.0 oC.

Mass of oil burned = 20.55 - 20.05 = 0.50 g

Temperature rise = 40.0 - 24.5 = 15.5oC.

Calculate the heat released in kg/g of vegetable oil.

Heat energy released = mass of water x heat capacity x temperature rise

Chemical potential energy released from the oil = 100  x  4.2 x 15.5 = 6510 J

Energy released from the oil = 6510 / 0.50 = 13020 J/g, 13.0 kJ/g (3 sf)

Although it is a better method than the boiling tube, the flame may be smoky and therefore incomplete combustion, so the calorific value of J/kg will be less than the theoretical 100% conversion of chemical potential energy.

There are still heat losses by convection up the side of the copper calorimeter.

WHAT NEXT?

TOP OF PAGE

INDEX of all my BIOLOGY NOTES

This is a BIG website, so try using the [SEARCH BOX], it maybe quicker than the many indexes!

for KS3 science students aged ~11-14, ~US grades 6, 7 and 8

ChemistryPhysics UK GCSE/IGCSE students age ~14-16, ~US grades 9-10

for pre-university ~16-18 ~US grades 11-12, K12 Honors