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Surface energy

Read about surface energy, Mechanical Properties of Fluids. Learn about low and high surface energy, surface energy formula, and much more.

Introduction 

This article will discuss the concept of surface energy from NCERT chapter 10-Mechanical properties of Fluids. We will discuss the concept of surface energy, its relation to surface tension, the surface energy formula, along with some solved examples. 

Defining surface energy 

  • When the surface of the liquid molecules exhibits excess energy compared to the molecules on the inside of the liquid, then that excess energy is called surface energy
  • Surface energy is described as potential energy in liquid molecules that help the molecules reach the surface of the liquid
  • Metals have high surface energies and perform well under different temperatures and environmental conditions. Metals are easy to wet out and come in contact with liquids. Therefore, it is easy to bond metals with most adhesives
  • Most plastics have low surface energy. These materials repel water and are difficult to bond together. However, engineered plastics are lightweight, low-cost, strong, and have higher surface energies
  • When two fluids are in contact, the surface energy will depend upon the materials of the fluid surfaces in contact. If the molecules of two fluids attract, their surface energy decreases. However, if the molecules of the two liquids in contact repel each other, then their surface energy increases

High surface energy- In materials that have high surface energy, the molecules on the surface are strongly attracted to each other. Due to the high attractiveness of their molecules, they are also easily attracted to liquid molecules. High surface energy materials have surface energies that range between 100s and 1000s dynes/cm. These materials are easy to wet out and bond.

Medium surface energy- The surface energies of medium surface energy range between 36 dynes/cm and 300 dynes/cm. Natural materials such as wood and stone, along with specifically engineered plastics and concrete, have medium surface energies.  

Low surface energy- Materials that have surface energy below 36 dynes/cm are considered low surface energy materials. The molecules of materials that have low surface energy have very little attraction between them, and they are quite happy the way they are. 

Surface energy and surface tension 

Relation between surface energy and surface tension is- 

Let us assume that a rectangular frame PORS has a movable wire CD and that QR = CD = L. If CQRD develops a layer of soap film, then the surface tension will try to pull the wire, CD, inwards by a force F. 

Surface tension = force free length

Force (f) = surface tension free length 

∴ F = T (2L)… (1)

If the surface tension will try to pull the wire, CD inwards by force F and distance ‘dx,’ then. 

Work done = F dx

By substituting the value of F in equation (1), we get 

Work done = T 2Ldx

However, increase in area = d A = 2Ldx

Therefore, using isothermal conditions, surface energy will be defined as the work done per unit area to increase the free surface area, 

Therefore, Surface energy = Work Done Free surface area = Work DonedA = T 2Ldx2Ldx

Surface energy formula 

Surface energy is mathematically represented as- 

Surface energy = Work DoneArea Joules/m2

Therefore, the SI unit of surface energy is Newton/meter or Joules/m2.  

Example – Let the surface tension of water be 72 10-3 N/m, and the length of the two parallel lines be 20 cm each. If the separation between the two parallel wires is 1cm, then what is the work required to increase the surface area by 2mm. 

Solution – From the statement mentioned above, we can gather the following information- 

The length of the two wires is 20 cm each. 

The separation between the two wires is 1cm = 10-2 m 10-3 m. 

The surface tension of the water = 72 10-3

Increase in the surface area = 2mm = 2  

As we have established in an earlier segment, increase in surface energy = T increase in the surface area. 

Increase in surface area = 2 20 2 10-3 = 80 × 10-3 m2 

Increase in surface energy = (72 10-3) (80 × 10-3) = 5.76 10-3 J 

Therefore, the required work to be done is 5.75mJ in order to increase the surface area by 2mm. 

Conclusion 

With this, we conclude our article on surface energy. We hope that we were able to clarify and add to your knowledge about surface energy.