Survival of Warm-Blooded Aminals in Arctic Conditions

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Aim:

To find out how warm-blooded animals survive in the Arctic conditions.

Background information:

The Arctic conditions are extreme; they consist of sub-zero temperatures, heavy snow, strong winds and ice. To survive these conditions animals must keep a constant body temperature to maintain the optimum temperature for essential chemical reactions to take place. If a living cell cools to a temperature below 0°C then ice crystals can form which will puncture the cell membrane causing the cell to burst. If a living cell reaches temperatures above 45°C then the proteins (including enzymes) are denatured; therefore no longer work. This means that the thermal limits for life are 0-45°C. Animals must regulate their body temperatures' between these bounds to survive. Mammals and birds stay closer to the upper bound.

Respiration

Glucose + oxygen Carbon dioxide + water + energy

C6H12O6 + 6O2 6CO2 + 6H2O + energy

Animals lose heat more rapidly in colder conditions. This is because heat moves from hot regions to cold so the colder the environment the quicker the animal will lose heat. Heat can only move by conduction, convection and radiation.

Animal maintain their temperatures in many ways. There are physical and behavioural ways of regulating temperature. The physical methods include;

Growing thick feathers/fur- therefore stopping heat loss by convection as fur and feathers trap air which then acts as an insulator, fat/blubber which as an insulator, oily or greasy skin which would make the animal waterproof and stops water from entering layers of air therefore maintaining insulation layers, dark skin to absorb more light and heat (black absorbs light, white reflects light, th...

... middle of paper ...

...28 23.5

5 18 22 27 22.3

17.5 22 25 21.5

5 17 21.5 23.5 20.7

17 21 23 20.3

My results for this beaker are mixed. There is a big difference between beaker 1 and beaker 3. Beaker 3 is 9°C warmer than beaker 1 at 4.5 minutes and has a difference of 6°C at 7 minutes. This would strongly affect my average. As there is such a difference between both of these beakers, I would not repeat them both as they both cant be wrong with such a difference. Therefore I would only repeat beaker 3 because beaker 1's readings are close to beaker 2's readings and beaker 3's reading are separate to both sets of results and has affected the average. By doing this I may affect the total decrease of this experiment which seems to follow a slight pattern between the beakers with insulation layers of 1,2 and 3.

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