If-a-number-of-little-droplets-all-of-the-same-radius-r-coalesce-to-form-a-single-drop-of-radius-R-show-that-the-rise-in-temperature-is-given-by-3T-pJ-1-r-1-R-where-T-is-surface-tension

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Question Number 51325 by peter frank last updated on 26/Dec/18

$$Ifanumberoflittle$$$$dropletsallofthesame$$$$radiusrcoalesceto$$$$formasingle$$$$dropofradiusR.show$$$$thattheriseintemperature$$$$isgivenby$$$$\frac{3T}{pJ}(\frac{1}{r}-\frac{1}{R})$$$$whereTissurfacetension$$$$ofwaterandJismechanical$$$$equivalentofheat$$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18

$$densityofliquid=\rho$$$$n\times \frac{4}{3}\pi r^3\times \rho =\frac{4}{3}\times \pi R^3\times \rho$$$$R^3={nr}^{3}n=\frac{R^3}{r^3}$$$$surfaceareaofndrops=n\times 4\pi r^2$$$$surfaceareaofbigdrop=4\pi R^2$$$$changeinsurfacearea=4\pi ({nr}^2-R^2)=4\pi (\frac{R^3}{r^3}{r}^2-R^2)$$$$=4\pi R^2(\frac{R}{r}-1)$$$$work=J\times m\times s\times △\theta s=specificheat△\theta =temparaturedifference$$$$work=surfacetension\times changeofsurfacearea$$$$w=T\times 4\pi R^2(\frac{R}{r}-1)$$$$J\left(\frac{4}{3}\pi R^3\times \rho \right)\times s\times △\theta =4\pi R^2\left(\frac{R-r}{r}\right)\times T$$$$J\times \rho \times s△\theta =3\left(\frac{R-r}{rR}\right)\times T$$$$△\theta =\frac{3T}{J\rho s}(\frac{1}{r}-\frac{1}{R})=\frac{3T}{J\rho }(\frac{1}{r}-\frac{1}{R})[s=1]$$$$$$

Commented by peter frank last updated on 26/Dec/18

$$thankyousir$$

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18

$$mostwelcome\dots$$

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