Surface-area-to-volume ratio

Surface Area to Volume Ratio and the Relation to the Rate of Diffusion Aim and Background This is an experiment to examine how the Surface Area / Volume Ratio affects the rate of diffusion and how this relates to the size and shape of living organisms. The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen

celled organisms are small meaning that their surface area is big compared to their volume. These organisms have a large surface area to volume ratio which means that they can obtain different substances by diffusion through their relatively large plasma membrane. The substances have to diffuse only short distances so they can diffuse at a faster rate and meet the organism’s needs. Multi-cellular organisms have a much smaller surface area to volume ratio. Many of their cells are not in direct contact

Surface area / Volume ratio Experiment Introduction: The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into

smaller surface area to volume ratio. Many of their cells are not in contact with their surroundings so they con not only rely on diffusion to supply all their organs with oxygen and nutrients, as the distance from their surface to all cells is too fare. We are multi cellular and have special surface for gaseous exchange and for obtaining nutrients. However Single celled organisms are small, which means that their surface area is large compared with their volume; they have a large surface area to volume

[Type text] [Type text] [Type text] _An experiment on the effect of surface area to volume ratio on the rate of osmosis of Solanum tuberosum L._ BACKGROUND A cell needs to perform diffusion in order to survive. Substances, including water, ions, and molecules that are required for cellular activities, can enter and leave cells by a passive process such as diffusion. Diffusion is random movement of molecules in a net direction from a region of higher concentration to a region of lower concentration

water in order to prepare concrete, the hardened concrete formed after hydration is not a solid mass but a porous material. There is formation of pores due to water which is initially added. Water occupies some fraction of the total initial volume and after the hydration of cement water is held physically, chemically or is adsorbed on product after hydration. This leads to formation of different types of pore structure inside the hydrated cement. CLASSIFICATION OF PORES STRUCTURES IN

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

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

relationship between surface area : volume ratio and heat loss. INTRODUCTION: The aim of this experiment is to investigate and find the relationship between heat loss (of water) and surface area to volume ratio of animals. To investigate this, we are going to use three flasks of different volume (as the equivalent the animals) and thus different surface areas filled with water. BACKGROUND: Surface Area : Volume Ratios We will be using the following formula for calculating SA:Vol ratios: SA : Vol Vol

1. Shape Surface Area Formula Volume Formula (cube 2 S.A. 6  s 3 V s  sphere 2 S.A. 4  r 4 3 3 V r   cylinder 2 S.A. 2 2     r Rh 2 V r h. Cube: s/6 Sphere: 0.333333r Cylinder: 6.283185h+6.283185r Volume to surface ratio: = s^3 / 6s^2 = s/6. Cube shaped sphere shaped cylinder shaped Volume 16 16 16 Surface area 38.097625 30.706532 35.207969 Volume/S.A Ratio .419973 .521061