# How Does the Surface Area to Volume Ratio Affect Heat Loss in Organisms?

Topics: Thermodynamics, Temperature, Volume Pages: 16 (4075 words) Published: December 2, 2012
Biology Lab Report
BY:
Michael Ryan Pranata 11C
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Background:
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 can be more “paths” from the sides of the body that are capable of releasing this heat particles, and reaching thermal equilibrium faster. This is what happens when a hotter body is subjected to a colder one. Research Question:

How does the surface area to volume ratio affect heat loss in organisms? Hypothesis:
I hypothesize that the larger the surface area to volume ratio, the more heat will be lost and vice versa. In this experiment, there will be a series of sizes of round bottom flasks, each having a different surface area to volume ratio. The smallest round bottom flask will have the biggest surface area to volume ratio, therefore the heat loss in that particular flask will be most, whereas the biggest flask will have the least surface area to volume ratio, therefore heat loss will be the least. The reason behind the difference in rate of heat loss can be explained by the fact that the surface area to volume ratio indicates the surface area, a particular unit volume has. This, in other words, states that in comparison to the content, the surface in contact with the outside is comparatively greater. Therefore, this can also be interpreted in a way to explain the rate of heat loss. Since more particles of water are able to contact the outer atmosphere, more heat is lost simultaneously, which results in a faster heat loss. Variables:

Independent: Surface area to volume ratio
On the basis of this experiment, this is the independent variable because the experiment desires to see the impact of the change of this variable to the heat loss. We change this variable by using different sizes of round bottom flasks. We’ve used four different surface area to volume ratio to observe the change in heat loss.

Surface Area to Volume Ratio|
1.132| 1|
0.882| 1|
0.652| 1|
0.517| 1|

A volume of round bottom flask can be calculated by the formula: Surface area: 4πr2
Volume: 43 πr3

*FOR ALL THE FLASKS, THE COLLECTION OF DIAMETER CAN BE SHOWN BY THE METHOD IN THE DIAGRAM BELOW*

* Diameter: 5.3 cm
* Volume: 43 πr3
= 43 x π x (5.32)3
= 77.592 cm3
* Surface Area: 4πr2
= 4 x π x (5.32)2
= 88.247 cm2
* Diameter: 6.8 cm
* Volume: 43 πr3
= 43 x π x (6.82)3
= 164.636 cm3
* Surface Area: 4πr2
= 4 x π x (6.82)2
= 145.267 cm2
* Diameter: 9.2 cm
* Volume: 43 πr3
= 43 x π x (9.22)3
= 407.720 cm3
* Surface Area: 4πr2
= 4 x π x (9.22)2
* = 265.904 cm2
* Diameter: 11.6 cm
* Volume: 43 πr3
= 43 x π x (11.62)3
= 817.283 cm3
* Surface Area: 4πr2
= 4 x π x (11.62)2
* = 422.733 cm2

Surface Area to Volume Ratio
Since the volume (whatever it may be for each flask) will be taken as unit (1), the surface area will be matched according to the volume. This is done by: Surface AreaVolume
Surface Area: 88.247 cm2
Volume: 77.952 cm3
Surface Area : Volume
88.24777.952 = 1.132
Therefore the ratio is 1.132 : 1

Surface Area: 145.267 cm2
Volume: 164.636 cm3
Surface Area : Volume
145.267164.636 = 0.882
Therefore the ratio is 0.882: 1
Surface Area: 265.904 cm2
Volume: 407.720 cm3
Surface Area : Volume
265.904407.720 = 0.652
Therefore the ratio is 0.652 : 1