Question Number 51325 by peter frank last updated on 26/Dec/18
$${If}\:{a}\:{number}\:{of}\:{little}\: \\ $$$${droplets}\:{all}\:{of}\:{the}\:{same}\: \\ $$$${radius}\:{r}\:{coalesce}\:{to} \\ $$$${form}\:\:{a}\:{single} \\ $$$${drop}\:{of}\:{radius}\:{R}.{show} \\ $$$${that}\:{the}\:{rise}\:{in}\:{temperature} \\ $$$${is}\:{given}\:{by} \\ $$$$\frac{\mathrm{3}{T}}{{pJ}}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{R}}\right) \\ $$$${where}\:\:{T}\:{is}\:{surface}\:{tension} \\ $$$${of}\:{water}\:{and}\:{J}\:{is}\:{mechanical} \\ $$$${equivalent}\:{of}\:{heat} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18
![density of liquid=ρ n×(4/3)πr^3 ×ρ=(4/3)×πR^3 ×ρ R^3 =nr^3 n=(R^3 /r^3 ) surface area of n drops=n×4πr^2 surface area of big drop=4πR^2 change in surface area=4π(nr^2 −R^2 ) =4π((R^3 /r^3 )r^2 −R^2 ) =4πR^2 ((R/r)−1) work=J×m×s×△θ s=specific heat △θ=temparature difference work=surface tension×change of surface area w=T×4πR^2 ((R/r)−1) J((4/3)πR^3 ×ρ)×s×△θ=4πR^2 (((R−r)/r))×T J×ρ×s△θ=3(((R−r)/(rR)))×T △θ=((3T)/(Jρs))((1/r)−(1/R))=((3T)/(Jρ))((1/r)−(1/R)) [ s=1]](https://www.tinkutara.com/question/Q51329.png)
$${density}\:{of}\:{liquid}=\rho \\ $$$${n}×\frac{\mathrm{4}}{\mathrm{3}}\pi{r}^{\mathrm{3}} ×\rho=\frac{\mathrm{4}}{\mathrm{3}}×\pi{R}^{\mathrm{3}} \:×\rho \\ $$$${R}^{\mathrm{3}} ={nr}^{\mathrm{3}} \:\:\:{n}=\frac{{R}^{\mathrm{3}} }{{r}^{\mathrm{3}} } \\ $$$${surface}\:{area}\:{of}\:{n}\:{drops}={n}×\mathrm{4}\pi{r}^{\mathrm{2}} \\ $$$${surface}\:{area}\:{of}\:{big}\:{drop}=\mathrm{4}\pi{R}^{\mathrm{2}} \\ $$$${change}\:{in}\:{surface}\:{area}=\mathrm{4}\pi\left({nr}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)\:=\mathrm{4}\pi\left(\frac{{R}^{\mathrm{3}} }{{r}^{\mathrm{3}} }{r}^{\mathrm{2}} −{R}^{\mathrm{2}} \right) \\ $$$$=\mathrm{4}\pi{R}^{\mathrm{2}} \left(\frac{{R}}{{r}}−\mathrm{1}\right) \\ $$$${work}={J}×{m}×{s}×\bigtriangleup\theta\:\:\:{s}={specific}\:{heat}\:\:\bigtriangleup\theta={temparature}\:{difference} \\ $$$${work}={surface}\:{tension}×{change}\:{of}\:{surface}\:{area} \\ $$$${w}={T}×\mathrm{4}\pi{R}^{\mathrm{2}} \left(\frac{{R}}{{r}}−\mathrm{1}\right) \\ $$$${J}\left(\frac{\mathrm{4}}{\mathrm{3}}\pi{R}^{\mathrm{3}} ×\rho\right)×{s}×\bigtriangleup\theta=\mathrm{4}\pi{R}^{\mathrm{2}} \left(\frac{{R}−{r}}{{r}}\right)×{T} \\ $$$${J}×\rho×{s}\bigtriangleup\theta=\mathrm{3}\left(\frac{{R}−{r}}{{rR}}\right)×{T} \\ $$$$\bigtriangleup\theta=\frac{\mathrm{3}{T}}{{J}\rho{s}}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{R}}\right)=\frac{\mathrm{3}{T}}{{J}\rho}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{R}}\right)\:\:\:\:\left[\:{s}=\mathrm{1}\right] \\ $$$$ \\ $$
Commented by peter frank last updated on 26/Dec/18
$${thank}\:{you}\:{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18
$${most}\:{welcome}… \\ $$