Find out the percentage of citric acid present in lemon squash by using a neutralisation method of titration or volumetric analysis, and by using sodium hydroxide as the neutralising agent (it is an alkali).

As the thermometers provided by the school are long and two of the beakers are too small to hold a thermometer up right I will use a clamp stand to hold up, this will reduce the chances of any apparatus breaking and any injuries.

Variables

For my experiment the parts will remain the same through out the investigation:

1. Similar thermometer (0oC- 110oC). This has to be the same because each thermometer could give slightly varying results.
2. Amount of water in each beaker (in the 500ml beaker I will use 500ml, in the 200ml beaker I will use 200ml,and in the 30ml beaker I will use 30ml). This has to be the same in each size as the more water there is the longer it will take to cool down as the volume increases, causing a change in the rate in which the gas will flow out.

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3. The glassware (the beakers for measuring the water, the beakers with the test, the thermometer, etc).
4. Beakers- 500ml, 200ml, and 30ml.
5. Beakers for collecting the hot water- 500ml X 2
6. Stopwatch- minutes & seconds
7. Person recording results- this person will have a different speed of recording and checking the time.
8. Same table of results- this is for repeats.

And the following could change

1. The temperature of the water- this could differ in each repeated experiment without you expecting it. For example in one investigation you start with 780C and in another you start with 620C this would not help when making averages. This could be a variable that you could investigate but it would be hard getting the same temperatures.
2. Thickness- these beaker are glass, by changing the thickness of the beakers (by adding and insulation) you could investigate how this affect the readings and would be able to conduct an investigation on another variable.

         Insulation

        Small beaker        

3. Size of the beaker- this could change if you wanted to investigate how the surface area affects the amount of heat lost by the water.

What apparatus will I need?

1. Beaker- This can be used because it has a large volume and it would be easier to try an investigation with out having it going wrong (the water could easily be measured). It will hold the hot water to cool and it could also act as the “skin” for the heat to pass through. There will be three different types of beakers that I will use; they are 500ml, 200ml, and 30ml.
2. Thermometer- This will be used to measure the temperature at timed intervals. This will give me a chance to plot a set of tabled and graph results. And will be constantly used to check how much the temperature has dropped. I will be using a 1000C thermometer.
3. Stop Clock- This will be used so that my time interval of 30 seconds which is needed for getting accurate results. The stop clock should measure minutes and seconds.
4. Clamp Stand- The clamp stand will serve as a safety feature, stopping the thermometer from falling, and will help in reading during a fair test (It will hold the thermometer in mid-water.

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5. Heatproof mat- This will simply stop the extreme heat of the water from melting the thin layer of varnish on the wood. It will also serve as a suspender, stopping any heat leaving the water other than through the air.
6. Gloves- This will be used as I have described in Safety.

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Evaporation

Evaporation of seat (from sweat ducts) has a cooling effect because heat energy (the ‘latent heat of vaporisation’) from the organism is used; this is the energy used to convert liquid water in sweat into vapour. The rate of evaporation is affected by humidity and other factors in the environment.

The rate of sweating increases when the body temperature begins to rise, e.g. because of exercise, or because the surroundings are relatively warm. An increase in the amount of sweat produced corresponds to an increase in the blood supply to the skin surface. Note that sweating is also a form of excretion, since the sweat contains urea and also excess minerals. Sweating obviously has an effect on the body’s water content (osmoregulation).

 Radiation

Radiation involves the direct transfer of heat energy from the blood to the external environment. This is affected by the amount of blood which enters fine blood vessels nearest the surface:

• Vasodilatation increases the loss of heat by radiation. Vasodilatation involves a widening of arterioles (narrow arteries) leading to the surface.
• Vasoconstriction reduces the loss of heat by radiation. Vasoconstriction involves a narrowing of arterioles leading to the surface.

Excessive heat loss from the body during cold conditions can lead to hypothermia. This is a serious risk for old people, who may be relatively inactive and who may have difficulty in affording sufficient food, clothing and fuel. Hypothermia is also a risk for those in cold water, since water rapidly conducts heat away from the body.

Surface area/ volume ratio

The relationship between surface area and the size and shape of organisms (or cells) is very important in CHEMISTRY. This is mainly because it is through the surfaces that substances are exchanged between the internal and external environment of an organism (or cell). Examples of this include gas exchange and temperature regulation.

Surface area and volume (i.e. size) can be compared as a ratio. The surface area/ volume ratio (s.a./vol.) can be written:

Surface area    or s.a.

                                                         Volume          vol.

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Although organisms (or cells) do not usually have a very regular, ‘geometric’ shape, surface area/ volume ratios can most easily be compared in cube-type shapes. (In the examples that follow, all the dimensions are of the same units.)

Effect of the size on the s.a./vol. ratio

Compare the cubes below. Cube 1 has a relatively large s.a. vol. Ratio (=6), while the ratio for cube 2 is small (=3). This means that substances would be exchanged more readily to and from cube 1 than cube 2. For example, cube 1 would tend to gain or lose heat more rapidly than cube 2. The relative advantages of this depend on the organism and its situation. For example, it might be an advantage to be a small lizard on a warm day, but a disadvantage to be a small baby on a cold day.

Cube 1

1                      1                

                                              2                                            2

       

        1

                                             

                                                                2

s.a. = 1x1x6=6                                    s.a.=2x2x6=24

vol.= 1x1x1=1                                    vol.= 2x2x2=8  

s.a.= 6 = 6                                           s.a.= 24 = 3          

vol.   1                                                 vol.   8

The effect of size on the surface area

Effect of shape on the s.a./vol. Ratio

Now compare cube 2 in the above diagram with cuboid 3 below. Note that these two solids have the same volume; they are different only in shape. The surface area of cuboid 3 is greater because of its ‘flattened’ shape; it therefore has a larger s.a./vol. Ratio (=4.25) than cube 2 and substances would be exchanged more readily with it.

                                           1

                                         

                                          1

8

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s.a.= 34

s.a = 34 =4.25 

vol.    8

Repeats

The reason why you should repeat the experiment a number of times are so that the result would then look more realistic. You would be able to see errors in the reading of experiment. I have decided to do this experiment three times and then work out an average reading for every 30 seconds and thus my results will be more accurate.

 

Number of readings

You have to do a number of repeats and the gain an average of the results. I will do the same experiment three times and have three different sets of results for every minute.

Preliminary experiment

Aim- The aim of the preliminary experiment is to check that the smaller beaker begins to cool most rapidly out of the three different sizes. Determining whether the apparatus works and my prediction works.

Prediction- My prediction for this experiment was that the smallest beaker will cool down most and also the fastest out of the three beakers, and when a graph is produced the big beaker line will be the flattest, and the one with the smallest beaker will have the steepest.

Apparatus- 1.  Beakers (500ml, 200ml, 30ml)

               2.  Hot Water

                   3.  Kettle

               4.  Heatproof mat

                5.  Thermometer

               6.  Stop clock

       

Method

First I collected the apparatus (mentioned above). And I then set up the apparatus, the beakers sat on heatproof mats, the thermometers were held in the middle of each beaker by a clamp stand on each beaker (look at the diagram on the next page if you do not follow). It then boiled a litre of water in the kettle and then poured it into a few large 500ml beakers. Then I made sure that my apparatus was set up and ready to start with. I then poured the water into each of the three beakers. As soon as the water was poured to the correct height (there were a few problems as there were unequal amounts and I had to keep on pouring again) I started the stop clock. I took down the

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results every thirty seconds like in the table below, I took a total set of results for ten minutes.

Diagram

Below is how my experiment way laid out as.

A cross section of how the apparatus will be set

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Results

Below are the results I got from the experiment I did with the water in no particular volume just an average fill.

On the next page is a graph that I made showing the rate of temperature drop against time.

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Calculations 

I decided to work out the gradient of the graph for the last two minutes of the graph for the beaker size of 30cm3

Rate of temperature drop = y axis

                                            X-axis

                           

                                        =50-48

                                             2

                                        =2

                                          2

                                        = 10C/min

The amount of heat being lost during the last 2 minutes was 10C/min

These results show-

These results show that the rate in which the heat is lost relatively stays the same over the period of ten minutes this could show the heat loss stays constant and does not fluctuate (that is if in a body it is at rest).

Conclusion-

From analysing the graph I found a possible misread and so I did an estimate and placed/ traced the increase with a red dashed line (on graph).

From looking at my results I can say that I have proved my hypothesis/prediction for the preliminary experiment. I predicted that the smallest beaker will cool down most and also the fastest out of the three beakers, and when a graph is produced the big beaker line will be the flattest, and the one with the smallest beaker will have the steepest. This was proven correct because by the end of ten minutes the temperatures on the thermometer had drop about 10-200C.

Investigation (future tense)

Aim & Prediction-

My aim for this experiment is to find if the smallest beaker will loose more heat and if the smaller one looses its heat quickest out of the three. And I predict that the smaller beaker the faster the heat loss. This I predicted because the surface area: volume ratio decreases, as the size of the beaker gets smaller.

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Method-

i) I am going to do an experiment to find out whether prediction is proved or not.

Firstly I will set up the apparatus as show on page 9. I will then collect one litre of boiling water from the kettle.

ii) I will make final checks on the apparatus, and as soon as every one is in place I will pour the water into the beakers and begin the timer. Then will record the results on a table, shown later.

iii) After that is done I will set out the results on a graph.

iv) Then if it is possible I will repeat the investigation, but this time the smallest beaker will have a fur insulation

Diagram-

i)

ii)

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Results-

i)

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ii)

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iii)

Calculation to be done-

I will do the following calculations-

The rate at which temperature drops= Temperature 1- temperature 2

                               Time 1- time 2

To work out the ratio= Surface area: volume ratio

                                     Surface area of beaker: the volume of hot water it is holding

                          And then simplify

                          This will give the surface area: volume ratio

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Conclusion-

I think that my predicted results will be close to what I will actually get if I were to do the experiment.