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Question: Calculate the amount of energy needed to break a drop of water \[2mm\] in diameter into \({10^9}\) d...

Calculate the amount of energy needed to break a drop of water 2mm2mm in diameter into 109{10^9} droplets of equal size, taking the surface tension of water 7.3×10−2Nm−17.3 \times {10^{ - 2}}N{m^{ - 1}}.

Explanation

Solution

Hint We will calculate the energy difference between the final and the initial drops and this difference will be the required energy needed to break the single big drop into 109{10^9} droplets of equal size. The required energy of a single drop due to the surface tension is given by E=TAE = TA.

Complete Step by step solution
The radius of the drop is given. The initial area of the liquid drop is πr2\pi {r^2} where the diameter d=2mmd = 2mm. Therefore, the area is
πr2=π(1mm)2=π(1×10−3m)2=π×10−6m2\pi {r^2} = \pi {(1mm)^2} = \pi {(1 \times {10^{ - 3}}m)^2} = \pi \times {10^{ - 6}}{m^2}.
Now the drop of water is made to break into 109{10^9} droplets of equal size. This process would keep the volume of water as constant.
43πrinitial3=109×43πrfinal3\dfrac{4}{3}\pi {r_{initial}}^3 = {10^9} \times \dfrac{4}{3}\pi {r_{final}}^3
Let rfinal{r_{final}} be RR, so that the above equation becomes r3=109R3{r^3} = {10^9}{R^3}.
⇒r=103R\Rightarrow r = {10^3}R, or
⇒R=10−3r=10−3(1mm)\Rightarrow R = {10^{ - 3}}r = {10^{ - 3}}(1mm)
∴R=10−6m\therefore R = {10^{ - 6}}m
The final surface area of all the droplets combined is 109×πR2{10^9} \times \pi {R^2}.
∴109×π(10−6m)2=π×10−3m2\therefore {10^9} \times \pi {({10^{ - 6}}m)^2} = \pi \times {10^{ - 3}}{m^2}.
We see that the surface area has increased considerably from its initial value when there was only one big drop.
Now the energy EE, stored in a drop due to the surface tension is given as
E=TAE = TA,
where AA is the surface area of the drop and TT is its surface tension.
Applying this formula in the initial and final situations, we get,
Einitial=TAinitial{E_{initial}} = T{A_{initial}} and Efinal=TAfinal{E_{final}} = T{A_{final}}.
⇒Einitial=TAinitial=(7.3×10−2Nm−1)(π×10−6m2)=7.3π×10−8Nm\Rightarrow {E_{initial}} = T{A_{initial}} = (7.3 \times {10^{ - 2}}N{m^{ - 1}})(\pi \times {10^{ - 6}}{m^2}) = 7.3\pi \times {10^{ - 8}}Nm
And,
⇒Efinal=TAfinal=(7.3×10−2Nm−1)(π×10−3m2)=7.3π×10−5Nm\Rightarrow {E_{final}} = T{A_{final}} = (7.3 \times {10^{ - 2}}N{m^{ - 1}})(\pi \times {10^{ - 3}}{m^2}) = 7.3\pi \times {10^{ - 5}}Nm
The required energy difference is the amount of energy needed to break the initial big drop into many small equal droplets.
This is found by Efinal−Einitial{E_{final}} - {E_{initial}}
⇒Efinal−Einitial=7.3π×10−5Nm−7.3π×10−8Nm=22.91×10−5Nm\Rightarrow {E_{final}} - {E_{initial}} = 7.3\pi \times {10^{ - 5}}Nm - 7.3\pi \times {10^{ - 8}}Nm = 22.91 \times {10^{ - 5}}Nm
∴ΔE=2.291×10−4Nm\therefore \Delta E = 2.291 \times {10^{ - 4}}Nm.
This is the extra energy required to break the drop.

Note We could have approached the problem in a different way. The energy of each individual drop of the final droplet formation could be calculated by the formula E=TAE = TA and then it could have been multiplied by the number of droplets which were finally formed, i.e. 109{10^9}. This would give the same required energy difference.