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Change in Surface Energy

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In this article. we shall study numerical problems on the change in surface energy due to breaking of drop or coalescence of number of drops into a one drop.

Change in Surface Energy

Example – 01:

Two mercury drops one of radius 2 mm and the other of radius 1 mm coalesce to form a single drop. Find the change in the free surface energy. S.T. of mercury = 0.544 N/m.

Given: radius of first drop = r= 2 mm = 2 × 10-3 m, radius of second bubble = r2 = 1 mm = 1 × 10-3 m, Surface tension = T = 0.544 N/m

To Find: Change in free surface energy = dU =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of two drops = Final volume of single drop

∴ 4/3 π r13 +    4/3 π r23 =     4/3 π r3

∴   r13 +   r23 =   r3

∴   23 +   13 =   r3

∴   r3 =   9

∴   r =   2.08 mm = 2.08 × 10-3 m

∴   Change in surface energy = dU = U1 – U2

∴    dU = T. A1 – T.A2   = T. (A1 – A2)

∴    dU = T. [(4πr1² +4πr2²)  – 4πr²) = T × 4π [(r1² +r2²)  – r²)]

∴    dU = 0.544 × 4π [((2 × 10-3)² + (1 × 10-3)²)  – (2.08 × 10-3)²)]

∴    dU = 0.544 × 4π [4 × 10-6 + 1 × 10-6 – 4.3264 × 10-6]

∴    dU = 0.544 × 4 × 3.142 × 0.6736 × 10-6

∴    dU = 4.61 × 10-6  J

Ans: The change in free surface energy is 4.61 × 10-6 J

Example – 02:

How much energy would be liberated if 103 water droplets each 10-8 m in diameter coalesce to produce a single drop? S.T. of water = 0.072 N/m. Assume drops are spherical.

Given: Initial number of drops = n1 = 103, initial diameter of each drop = d= 10-8 m, initial radius of each drop = r = 0.5 × 10-8 m = = 5 × 10-9 m Final number of drop = n2 = 1, Surface tension = T = 0.072 N/m

To Find: Energy liberated = dU =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n1 drops = Final volume of n2 drop

∴   n× 4/3 π r13   =   n× 4/3 π r3

∴   n1 ×   r13   =   n× r3

∴   103 ×   (5 × 10-9)3   =   1× r3

∴   10 × 5 × 10-9 =     r

∴ r = 5 × 10-8 m

∴   Energy released = dU = U1 – U2

∴    dU = T. A1 – T.A2   = T. (A1 – A2)

∴    dU = T. ( n × 4πr1²   – n × 4πr²) = T × 4π (n ×  r1²  –  n × r²)

∴    dU =  0.072 × 4π [103 ×  (5 × 10-9)²  –  1 × (5 × 10-8)²]

∴    dU =  0.072 × 4 × 3.142 × [103 × 25 × 10-18 –  25 ×10-16]

∴    dU =  0.072 × 4 × 3.142 × [250 × 10-16 –  25 ×10-16]

∴    dU =  0.072 × 4 × 3.142 × 225 × 10-14

∴    dU =  2.036 × 10-14 J

Ans: The energy liberated is 2.036× 10-14 J

Example – 03:

A drop of mercury of radius 0.1 cm is broken into 8 droplets of the same size. Find the work done if the surface tension of mercury is 540 dyne/cm.

Given: Initial number of drops = n1 = 1, initial radius of each drop = r= 0.1 cm = 0.1 × 10-2 m = 1 × 10-3 m, Final number of drop = n2 = 8, Surface tension = T =540 dyne/cm = 540 × 10-3 N/m

To Find: Work done = W =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n1 drops = Final volume of n2 drop

∴   n× 4/3 π r13   =   n× 4/3 π r3

∴   n×   r13   =   n× r3

∴   1 ×   (1 × 10-3)3   =   8× r3

∴ 1 × 10-3=     2 r

∴ r = 0.5 × 10-3 m

∴   Work done = Change in surface energy = dU = U1 – U2

∴    dU = T. A2 – T.A1   = T. (A2 – A1)

∴    dU = T. (n × 4πr²   –   n × 4πr1² ) = T × 4π (n × r²  –  n ×  r1² )

∴    dU =  540 × 10-3 × 4π [8  ×  (0.5 × 10-3)²  –  1 × (1 × 10-3)²]

∴    dU =  540 × 10-3 × 4π [8  ×  0.25 × 10-6 –  1 × 10-6]

∴    dU =  540 × 10-3 × 4π [2 × 10-6 –  1 × 10-6]

∴    dU =  540 × 10-3 × 4 × 3.142  × 1 × 10-6

∴    dU =  6.786 × 10-6 J

Ans: The work done is 6.786 × 10-6 J

Example – 04:

A mercury drop of radius 10-3 m breaks up into 125 small droplets. Calculate the change in energy assuming that the drops are spherical and S.T. of the mercury is 0.55 N/m.

Given: Initial number of drops = n1 = 1, initial radius of each drop = r= 10-3 m, Final number of drop = n2 = 125, Surface tension = T =0.55 N/m

To Find: Change in area = dU =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n1 drops = Final volume of n2 drop

∴   n× 4/3 π r13   =   n× 4/3 π r3

∴   n×   r13   =   n× r3

∴   1 ×   (10-3)3   =   125× r3

∴ 1 × 10-3=     5 r

∴ r = 0.2 × 10-3 m

∴   Work done = Change in surface energy = dU = U1 – U2

∴    dU = T. A2 – T.A1   = T. (A2 – A1)

∴    dU = T. (n × 4πr²   –   n × 4πr1² ) = T × 4π (n × r²  –  n ×  r1² )

∴    dU =  0.55 × 4π [125  ×  (0.2 × 10-3)²  –  1 × ( 10-3)²]

∴    dU =  0.55 × 4π [125  ×  0.04 × 10-6 –  1 × 10-6]

∴    dU =  0.55 × 4π [5 × 10-6 –  1 × 10-6]

∴    dU =  0.55 × 4 × 3.142  × 4 × 10-6

∴    dU =  2.76 × 10-5 J

∴    Change in energy is 2.76 × 10-5 J

Example – 05:

A mercury drop of radius 0.1 cm breaks up into 27 small droplets. Calculate the work done assuming that the drops are spherical and S.T. of the mercury is 540 dyne/cm.

Given: Initial number of drops = n1 = 1, initial radius of each drop = r= 0.1 cm = 0.1 × 10-2 m = 1 × 10-3 m, Final number of drop = n2 = 27, Surface tension = T =540 dyne/cm = 540 × 10-3 N/m

To Find: Work done = W =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n1 drops = Final volume of n2 drop

∴   n× 4/3 π r13   =   n× 4/3 π r3

∴   n×   r13   =   n× r3

∴   1 ×   (1 × 10-3)3   =   27× r3

∴ 1 × 10-3=     3 r

∴ r = 0.33 × 10-3 m

∴   Work done = Change in surface energy = dU = U1 – U2

∴    dU = T. A2 – T.A1   = T. (A2 – A1)

∴    dU= T. (n × 4πr²   –   n × 4πr1² ) = T × 4π (n × r²  –  n ×  r1² )

∴    dU =  540 × 10-3 × 4π [27  ×  (0.33 × 10-3)²  –  1 × (1 × 10-3)²]

∴    dU =  540 × 10-3 × 4π [27  ×  0.1089 × 10-6 –  1 × 10-6]

∴    dU=  540 × 10-3 × 4π [3 × 10-6 –  1 × 10-6]

∴    dU =  540 × 10-3 × 4 × 3.142  × 2 × 10-6

∴    dU =  1.357 × 10-5 J

Ans: The work done is 1.357 × 10-5 J

Example – 06:

Eight droplets of water each of radius 0.2 mm coalesce together to form a single drop, Find the change in total surface energy, given: surface tension of water = 0.072 N/m.

Given: Initial number of drops = n1 = 8, initial radius of each drop = r= 0.2 cm = 0.2 × 10-3 m = 2 × 10-4 m, Final number of drop = n2 = 1, Surface tension = T =0.072 N/m

To Find: Change in total surface energy = dU =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n1 drops = Final volume of n2 drop

∴   n× 4/3 π r13   =   n× 4/3 π r3

∴   n×   r13   =   n× r3

∴   8 ×   (2 × 10-4)3   =   1× r3

∴ 64× 10-12=     2 r

∴ r = 4 × 10-4 m

∴   Work done = Change in surface energy = dU = U1 – U2

∴    dU = T. A2 – T.A1   = T. (A1 – A2)

∴    dU = T. (n × 4πr1²   –   n × 4πr² ) = T × 4π (n × r²  –  n ×  r² )

∴    dU =  0.072 × 4π [8  ×  (2 × 10-4)²  –  1 × (4 × 10-4)²]

∴    dU =  0.072 × 4π [8  ×  4 × 10-8 –  16 × 10-8]

∴    dU =  0.072 × 4π [32 × 10-8 –  16 × 10-8]

∴    dU =  0.072 × 4 × 3.142  × 16 × 10-8

∴    dU =  1.448 × 10-7 J

Ans: The change in surface energy is 1.448 × 10-7 J

Example – 07:

A mercury drop of radius 0.5 cm falls from a height on glass plate and breaks up into million droplets, all of the same size, Find the height from which the drop must have fallen. Density of mercury 13600 kg/m3. Surface tension of mercury = 0.465 N/m

Given: Initial number of drops = n1 = 1, initial radius of each drop = r= 0.5 cm = 0.5 × 10-2 m = 5 × 10-3 m, Final number of drop = n2 = 106, Surface tension = T = 0.465 N/m

To Find: Height from which the drop falls = h =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n1 drops = Final volume of n2 drop

∴   n× 4/3 π r13   =   n× 4/3 π r3

∴   n×   r13   =   n× r3

∴   1 × (5 × 10-3)3   = 106× r3

∴ 5 × 10-3=     102 r

∴ r = 5 × 10-5 m

∴   Work done = Change in surface energy = dU = U1 – U2

∴    dU = T. A2 – T.A1   = T. (A2 – A1)

∴    dU= T. (n × 4πr²   –   n × 4πr1² ) = T × 4π (n × r²  –  n ×  r1² )

∴    dU=  0.465 × 4π [106 ×  (5 × 10-5)²  –  1 × (5 × 10-3)²]

∴    dU =  0.465 × 4π [106 × 25 × 10-10 –  25 × 10-6]

∴    dU =  0.465 × 4π [2500 × 10-6 –  25 × 10-6]

∴    dU=  0.465 × 4 × 3.142  × 2475 × 10-6

∴    dU =  0,01446J

∴    Work done is 0.01446J

Now, by work energy principle

Poential energy = work done

∴    mgh = 0.01446

∴    (Volume × density) gh = 0.01446

∴   4/3 π r× ρ × g ×h = 0.01446

∴   4/3 × 3.142 × (5 × 10-3)× 13600 × 9.8 ×h = 0.01446

∴   h = (3 × 0.01446) / (4 × 3.142 × (5 × 10-3)× 13600 × 9.8)

∴   h = (3 × 0.01446) / (4 × 3.142 × 125 × 10-9 × 13600 × 9.8)

∴   h = 0.2072 m

Ans: The height from which the drop may have fallen is 0.2072 m.

Example – 08:

n droplets of equal size of radius r coalesce to form a bigger drop of radius R. If T is the surface tension of the liquid, then show that the energy liberated is 4π R²T (n1/3   – 1)

Solution:

The initial volume of n drops = Final volume of one drop

∴   n×  4/3 π r3   =   1 ×  4/3 π R3

∴ n ×  r3  =   R3

∴ R = r × n1/3

∴ R × n– 1/3 = r

∴   Energy liberated = dU = U1 – U2

∴    dU = T. A1 – T.A2   = T. (A1 – A2)

∴    dU = T. ( n × 4πr²   – 1 × 4πR²)

∴    dU = T. ( n × 4π(R × n– 1/3)²   – 1 × 4πR²)

∴    dU = T. ( n × 4πR² × n– 2/3   – 1 × 4πR²)

∴    dU = T. 4πR² ( n × n– 2/3   – 1)

∴    dU = 4π R²T (n1/3   – 1)

Example – 09:

n droplets of equal size of radius r coalesce to form a bigger drop of radius R. If T is the surface tension of the liquid, then show that the energy liberated is 4πr² T. (n – n2/3)

Solution:

The initial volume of n drops = Final volume of one drop

∴   n×  4/3 π r3   =   1 ×  4/3 π R3

∴   n ×   r3   =   R3

∴ R = r × n1/3

∴   Energy liberated = dU = U1 – U2

∴    dU = T. A1 – T.A2   = T. (A1 – A2)

∴    dU = T. ( n × 4πr²   – 1 × 4πR²)

∴    dU = T. ( n × 4πr²   – 1 × 4π(r  × n1/3)²)

∴    dU = T. ( n × 4πr²   – 1 × 4π r²  × n2/3)

∴    dU = T. 4πr²( n  –  n2/3)

∴    dU = 4πr² T. ( n  –  n2/3)

Note: 

For competitive exams use following formulae directly

Coalescence of n droplets of radius r into a single drop of radius R, then energy liberated

dU = 4πr² T. ( n  –  n2/3) or dU = 4π R²T (n1/3   – 1)

Breaking of one drop of radius R into n droplets of radius r, then energy liberated

dU = 4πR² T. ( n  –  n2/3) or  dU = 4πr²T (n1/3   – 1)

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