Hypothesis Larger animals lose heat at a lower rate than smaller animals. As the organism gets bigger the volume and surface area ratio decreases. This means that there is less surface area to lose heat from. If we use penguins as an example, the larger penguins live closer to the poles that the smaller penguins.
So smaller animals have a higher surface area to volume ration and larger animals are the opposite. As the size of the animal increases the rate f heat loss decreases. The rate of heat loss depends on the surface area to volume ratio. The total heat production of ‘warm – blooded’ animals- depends upon the volume of metabolically active tissues whilst the rate of heat loss depends upon the surface area.
For this reason, animals living in colder climates tend to be large whilst living in a hot climate they are generally smaller. This is known as Bergmann’s rule and is observed in many species. The metabolic rate per gram of the smallest mammals is approximately 100 times faster than the largest. They have a large appetite enabling them to maintain a high metabolic rate. Heat gain and loss depend on the ratio of the surface area to the volume (the Surface Area/Volume ratio) because heat is stored in the volume of an animal but gained or lost over the surface. The higher the Surface Area/Volume, the faster an animal can heat or cool.
The Essay on Urban Heat Islands City Areas Pollutants
Urban Heat Islands For more than 100 years, it has been known that two adjacent cities are generally warmer than the surrounding areas. This region of city warmth, known as an urban heat island, can influence the concentration of air pollution. The urban heat island is formed when industrial and urban areas are developed and heat becomes more abundant. In rural areas, a large part of the incoming ...
Since as animals get larger, the Surface Area/Volume gets smaller (Volume increases faster than Surface Area as animals get larger), large animals heat or cool more slowly than do small animals. Length of One side Total surface Area Volume Ratio of surface area to volume 1 6 1 6: 1 2 24 8 24: 8 = 3: 1 3 54 27 54: 27 = 2: 1 4 96 64 96: 64 = 3: 2 = 1. 5: 1 5 125 125 150: 125 = 30: 25 = 1. 2: 1 6 216 612 216: 216 = 1: 1. radiation refers to light energy.
Animals can gain heat from sunlight; they can lose heat as heat energy radiated away from their bodies… conduction refers to gain or loss of heat from one surface to another. Animals can receive heat from warm surfaces or lose it to cold surfaces… convection refers to gain or loss of heat to or from a fluid, such as air or water. Wind or water currents often carry heat away from animals; occasionally they can also bring heat to animals… evaporative water loss leads to heat loss when water evaporates off of the body surface of an animal because it takes energy to convert water from liquid form to gas form; this energy is lost from the animal as heat.
Note that this only works if the water actually evaporates from the surface of the animal — if it drips off, heat is not lost. Method Equipment list. – Kettle – Beakers – Thermometer – Clamp stand and arm – Heat proof mat – Stopwatch – Water bath I will use 5 different size beakers to represent 5 animals at different sizes. The boiled water will be measured and put into the beakers and the temperature recorded. A stopwatch will be used to time 10 minutes and every minute the temperature will be taken. For each beaker of water the experiment will be done twice to ensure it is a far test and see if the results match closely.
The boiling water will be used immediately to prevent heat loss. My starting temperature will be 80 C. Variables Independent variable is the size of beaker and volume of water. I am using 5 different size beakers. Below shows some variables and how each of these influence my real = u lts also how I plan to control some of them. Dependent variable is the rate of heat loss.
The Essay on How Does the Surface Area to Volume Ratio Affect Heat Loss in Organisms?
As heat is a form of thermal energy, they tend to have the behavior of reaching a thermal equilibrium. This means that when two bodies of different temperatures come in contact with each other, the hotter ones will transfer heat particles to the body with a colder temperature, with an aim to reach this “thermal equilibrium”, whatever the temperature may be. The larger the surface area, means there ...
The control variables are to be kept constant to make it a fir test. e. g. The thickness of the glass beakers so there is equal insulation. This could effect my results because if a beaker was thinner than the others then of course the heat would escape faster.
The same starting temperature of hot water. If this was not the same for each experiment then the results would not prove anything. Use a thermometer. Constant ambient temperature.
Keep the difference between ‘body’ temperature and environment the same. Air movement to minimise heat loss by convection. Keep windows closed. Heat proof mat to keep surface or conduction constant. Otherwise the heat would escape through conduction. The results were as follows: Size of beaker Time/ minutes Temperature/ C 1 st time Repeat 1000 ml 012345678910 80 C 79 C 78 C 76 C 75 C 75 C 74 C 73 C 72 C 71 C 70 C 80 C 79 C 77 C 76 C 75 C 74 C 74 C 73 C 72 C 71 C 70 C 500 ml 12345678910 80 C 76 C 73 C 71 C 70 C 68 C 66 C 65 C 64 C 62 C 80 C 77 C 73 C 71 C 69 C 67 C 66 C 65 C 64 C 62 C 400 ml 12345678910 80 C 77 C 75 C 74 C 72 C 70 C 69 C 68 C 66 C 65 C 80 C 77 C 76 C 74 C 73 C 71 C 69 C 68 C 65 C 64 C 125 ml 12345678910 80 C 76 C 73 C 66 C 65 C 63 C 60 C 59 C 57 C 56 C 80 C 76 C 74 C 70 C 66 C 63 C 60 C 59 C 57 C 56 C 50 ml 12345678910 80 C 76 C 73 C 71 C 68 C 66 C 64 C 62 C 59 C 57 C 80 C 77 C 74 C 71 C 68 C 66 C 64 C 62 C 58 C 56 C Rate of heat loss results are as follows: Rate of heat loss = Total drop in temperature/Total time v 1000 ml Rate of heat loss = 10 C/10 min = 1 C/min v 500 ml Rate of heat loss = 18 C/10 min = 1.
8 C/min v 400 ml Rate of heat loss = 14 C/10 min = 1. 4 C/min v 125 ml Rate of heat loss = 24 C/10 min = 2. 4 C/min v 50 ml Rate of heat loss = 23 C/10 min = 2. 3 C/min I took all the measurements by having the thermometer in the water constantly and then every minute I took the temperature. I did two experiments at once to save time, and then I repeated them for a fair test.
