2. Convection - This is the transfer of heat by currents of air. Hot air rises and cooler air sinks, setting up convection currents around relatively hot objects. This is another means of the animal losing heat, but it is also minimised by prevention of these currents, for example by trapping air in a fibrous layer outside the skin. Fur in mammals and feathers in birds perform this function.
3. Radiation - This is energy that travels from one place to another as electromagnetic waves. Infrared radiation accounts for most of heat energy radiated and absorbed by animals.
So now that I know how heat is lost, I am going to have a more detailed look at the other side of the story, the size of the object.
I can almost definitely say that there is a connection between the size of the object and the heat lost. However, talking about, “Size,” in general is quite imprecise, as it is possible to be talking about either of two measures - surface area or volume. Both these factors are going to be relevant in this investigation as the greater the volume of the body, the more heat will be produced, and the greater the surface area of the body the more heat will be lost. A good example of this is a radiator - which has a large surface area so that more heat can be given out, or lost. So I am now searching for a link between these two factors, which could have a relationship with heat loss. I think that as we are comparing them, a ratio might be useful in these experiments. To investigate this idea further I will consider cubes, once again as hypothetical model animals.
CUBE 1
Side - 1cm
Surface area - 6cm2
Volume - 1cm3
Surface area to Volume ratio - 6 : 1
CUBE 2
Side - 4cm
Surface area - 96cm2
Volume - 64cm3
Surface area to Volume ratio - 1.5 : 1
CUBE 3
Side - 8cm
Surface area - 384cm2
Volume - 512cm3
Surface area to Volume ratio - 0.75 : 1
A hypothetical cube of side n cm :-
CUBE 4
Side - n
Surface area - 6n2
Volume - n3
Surface area to Volume ratio - 6/n : 1
So I have now discovered a method of linking surface area to volume in order to obtain a measure that can be used during the course of this investigation which is more precise than just size.
From looking at these cubes I can say that as the object increases in size, the surface area to volume ratio decreases.
I predict that the smallest flask I use will have the highest rate of heat-loss because it will have the largest surface area to volume ratio and that the largest flask I use will have the slowest rate if heat-loss because it will have the smallest surface area to volume ratio.
Scientific experiments need to be carried out with maximum precision if a ‘fair test’ is to be obtained. There are many factors which could have an effect on this investigation, amount of water used, shape of the flask used, thickness of the glass of the flask used, temperature of the surrounding air and size, or surface area to volume ratio, of the flask used to name but a few. However, for a ‘fair test’ to happen, only one of these factors can be changed at a time.
- The independent variable in this investigation, i.e. the one which changes will be the size if the flask. Sizes will range from 50millilitres to 5 litres.
- The dependent variable, i.e. what is being measured, will be heat loss, or more correctly, the temperature of the water at predetermined times.
- The controlled variables, i.e. the factors that will stay the same throughout the investigation are -
-
Thickness and shape of the flask - I can ensure these factors remain constant throughout the investigation because I will be using the same type of flask for each experiment, the only difference being the size of the flask.
-
Amount of water used - Although the actual amount of water used is really what is being changed, the amounts have to be relative to each other. Due to this I will fill the flask up to the same place each time. This place will be the bottom of the flask’s neck.
-
Surrounding air temperature - I will not be able to do much to control this, but I will carry the experiments out in the shortest space of time possible. This way there won’t be a great variance in the surrounding temperature.
On obtaining my results, I will record them in a table with the following headings: -
____ml Flask -
Heat loss - ___ oC / min
Adding the two temperatures recorded for any one time and dividing this total by two can work out the average temperature. By working out an average we are obtaining a more accurate result with less experimental error.
I will then process these raw results and obtain a rate of heat loss. This rate is measured per minute and can be worked out by using the following formula -
Heat loss = Highest Temperature - Lowest Temperature
Total Time Taken
I will then draw graphs of these results. The graphs will be: -
- Lines of average temperature verses time -
average
temp
( oC)
Time (mins)
- A bar chart of total heat loss against volume
of flask -
Total heat
loss ( oC)
Volume of flask (ml)
3.Lines of heat loss verses surface area : volume -
Rate of heat
loss
( oC / min)
Surface area : volume
After carrying out the experiments, recording the results, processing them and drawing graphs on the results, I will study the evidence I have obtained and draw conclusions from it. I will also discover if they prove my hypothesis that the larger flasks will retain more heat.
I will now carry out the experiments using the method shown on the next page.
METHOD
- Set up the apparatus as shown on the next page.
- Draw out a table with the headings shown before.
- Fill a kettle with water and heat until boiling.
- Pour this boiling water into the flask attached to the retort stand.
- Start the stop-clock when the reading on the thermometer is the same as your predetermined starting point.
- Measure the temperature each minute for 20 minutes, recording the results obtained in the table each time.
- Allow the flask to cool, and then repeat steps 1-6 using the same flask.
- Repeat steps 1-7 with different sizes of flasks.