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Surface energy is the energy in which the equivalent attractive force is present between the surface molecules of solid materials or substances. The energies differ from low to high or vice versa. Surface energy can not be measured. It arises when there is an interaction between molecules. The free surface of the liquid is called a stretched membrane. The surface stores some Potential Energy and it’s on the liquid surface, which is known as Surface Free Energy.
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What is Surface Energy?
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Surface energy is also known as interfacial free energy. It is the potential energy of the liquid molecules in which the molecules are the same at the surface of the liquid substance.
Surface energy decreases when the molecules at the liquid surface try to settle down at the bottom layer. The work is done to increase the surface area of the liquid surface. Equivalent attractive force presents between the surface molecules of solid materials or substances.
The mathematical expression of Surface Energy is
work done area joules/ m2.
The dimension of the surface energy is [MT-2] and the SI unit of surface energy is N/ m.
When the surface is larger, low surface energy will be exerted on the surface. When the surface is smaller, high surface energy will be exerted on the surface.
Examples of low surface energy are rubbers, plastics, etc. while oxides, ceramics, metals, etc. are examples of high surface energy.
The materials which have high surface energy are known as high surface energy materials.
Surface Energy of certain materials are given below:
| Materials | Surface Energy |
|---|---|
| Polystyrene | 40.6 |
| Water | 73 |
| Lead | 442 |
| Glass | 83.4 |
Relation Between Surface Energy and Surface Tension
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Let us take a rectangular wire framework. Dip the frame in and out of a soap solution. Two surfaces will be created due to soap film formation. The two surfaces are created in contact with the sliding wire. Due to the two surfaces, the surface tension acts on the wire.
Let the surface tension of the soap solution be T and the length of the wire be L.
The exerted force on each surface of the wire is T × L.
So, the total force acted on the wire = 2 T L.
Let the surfaces contract be
?x.Work done on the film be W=F ×
?x=2 T L
?xHere 2 L ×
?x is the total increase in the area of two surfaces of the film.W=T
? AOr T= W
?ASo, the work done is equal to the surface tension of the liquid in increasing the surface area of the free surface by 1 unit. So, the surface energy is equal to surface tension per unit area.
Es=T ?AThe unit of T is Jm-2 and the dimensional formula is [M1L0T-2]
Read Also: Reynolds Number
Surface Energy Formula
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Surface energy formula
surface area = work done area joules/ m2.
Joule = newton × m
Surface Tension of the liquid
σ=F/ L
Where it indicates, is the Surface tension of the liquid, F is the force per unit length and L is the line on which the force acts.
Read More: Bulk Modulus
Points to Remember
Read More: Young's Modulus
Sample Questions
Ques. Liquid medicine of density p is to be put in the eye. So, it is put in a dropper. When the bulb of the dropper is pressed, a drop forms on the tip. We estimate the size of the drop. The assumption is that the drop formed is spherical in shape. To determine the size of it, we calculate the net vertical force due to surface tension T and the radius is R. The drop detaches from the dropper when the force exerted becomes smaller than the weight of the drop.
If r=5 ×10-4 m, p=103 Kg/m3, T=0.11 N/m, g=10 m / s2, when it detaches from the dropper is approximate, what is the radius? (2 Marks)
1.4 × 10-3 m
1.4 × 10-3 m
1.4 × 10-3 m
1.4 × 10-3 m
Ans. When the weight of the dropper is equal to the vertical force due to the surface tension, then the drop will detach from the dropper that is,
43 π R3 pg=2 π r2T/ R
Now, solve the equation,
R=[ 3 r2 T/ (2 pg)]14≈1.4 × 10-3 m
Hence, the correct option is a.
Ques. A light square wireframe each side of which is 10 cm long hangs vertically in the water with one side just touching the water surface. Find the additional force necessary to pull the frame clear of the water ( T = 0.074 N/ m). (2 Marks)
Ans. Length of the frame which is in contact with the water = 1 = 10 cm = 10 × 10-2 m,
Surface tension T = 0.074 N/ m.
Now,
The water wets the wire from two sides, so, the effective length = 21
Also, T = F/ effective length
F = T × effective length
T × 21
F = 0.074 × 2 × 10 × 10-2 = 0.0148 N
Hence, the required pull is 0.0148 N.
Ques. A film of water is formed between two parallel wires, the length of each is 20 cm and the separation between them is 1 cm. to increase the surface area by 2 mm, find the work required. (Hint: surface tension of water is 72 × 10-3N/ m). (3 Marks)
Ans. Given: the length of a wire = 20 cm.
The separation between them = 1 cm or 10-2 m
Due to the external force, the displacement is = 2 mm = 2 10-3 m
The surface tension T = 72 × 10-3N/ m
Now, we know the work done, the relation between the surface energy and surface tension,
⇒?∪ =T. ? S …. (i)
? S=2 | ?x (increase in surface area)
? S=2×20 ×2 ×10-3=80 × 10-3 m2 ….. (ii)
Using equation (i) & (ii), we get
⇒? =72 × 10-3. 80 × 10-3=5.76 × 10-3
⇒?∪ =5.76 mJ
Hence, the work done to increase the surface area by 2 mm is 5.76 mJ.
Ques. The radius of a thin and light ring of the material is 3 cm and rests flat on the surface of the liquid. When slowly raised, it is found that the pull exerted on it is 0.03 N more before the film breaks than after. Find the surface tension. (2 Marks)
Ans. The radius of the ring material = 1 = 3 cm = 3 × 10-2 m, and the pull exerted is denoted with F = 0.03 N
The liquid wets the ring from two sides that is internal and external,
The effective length will be 2 × 2 π r
Now, T = F / effective length
T = F / (2 × 2 π r) = 0.03 / 2 × 2 × 3.142 × 3 × 10-2
= 0.08 N / m
Hence, the surface tension is 0.08 N / m.
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