Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

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  ((3T)/(pJ))((1/r)−(1/R))  where  T is surface tension  of water and J is mechanical  equivalent of heat

$${If}\:{a}\:{number}\:{of}\:{little}\: \\ $$$${droplets}\:{all}\:{of}\:{the}\:{same}\: \\ $$$${radius}\:{r}\:{coalesce}\:{to} \\ $$$${form}\:\:{a}\:{single} \\ $$$${drop}\:{of}\:{radius}\:{R}.{show} \\ $$$${that}\:{the}\:{rise}\:{in}\:{temperature} \\ $$$${is}\:{given}\:{by} \\ $$$$\frac{\mathrm{3}{T}}{{pJ}}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{R}}\right) \\ $$$${where}\:\:{T}\:{is}\:{surface}\:{tension} \\ $$$${of}\:{water}\:{and}\:{J}\:{is}\:{mechanical} \\ $$$${equivalent}\:{of}\:{heat} \\ $$

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]

$${density}\:{of}\:{liquid}=\rho \\ $$$${n}×\frac{\mathrm{4}}{\mathrm{3}}\pi{r}^{\mathrm{3}} ×\rho=\frac{\mathrm{4}}{\mathrm{3}}×\pi{R}^{\mathrm{3}} \:×\rho \\ $$$${R}^{\mathrm{3}} ={nr}^{\mathrm{3}} \:\:\:{n}=\frac{{R}^{\mathrm{3}} }{{r}^{\mathrm{3}} } \\ $$$${surface}\:{area}\:{of}\:{n}\:{drops}={n}×\mathrm{4}\pi{r}^{\mathrm{2}} \\ $$$${surface}\:{area}\:{of}\:{big}\:{drop}=\mathrm{4}\pi{R}^{\mathrm{2}} \\ $$$${change}\:{in}\:{surface}\:{area}=\mathrm{4}\pi\left({nr}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)\:=\mathrm{4}\pi\left(\frac{{R}^{\mathrm{3}} }{{r}^{\mathrm{3}} }{r}^{\mathrm{2}} −{R}^{\mathrm{2}} \right) \\ $$$$=\mathrm{4}\pi{R}^{\mathrm{2}} \left(\frac{{R}}{{r}}−\mathrm{1}\right) \\ $$$${work}={J}×{m}×{s}×\bigtriangleup\theta\:\:\:{s}={specific}\:{heat}\:\:\bigtriangleup\theta={temparature}\:{difference} \\ $$$${work}={surface}\:{tension}×{change}\:{of}\:{surface}\:{area} \\ $$$${w}={T}×\mathrm{4}\pi{R}^{\mathrm{2}} \left(\frac{{R}}{{r}}−\mathrm{1}\right) \\ $$$${J}\left(\frac{\mathrm{4}}{\mathrm{3}}\pi{R}^{\mathrm{3}} ×\rho\right)×{s}×\bigtriangleup\theta=\mathrm{4}\pi{R}^{\mathrm{2}} \left(\frac{{R}−{r}}{{r}}\right)×{T} \\ $$$${J}×\rho×{s}\bigtriangleup\theta=\mathrm{3}\left(\frac{{R}−{r}}{{rR}}\right)×{T} \\ $$$$\bigtriangleup\theta=\frac{\mathrm{3}{T}}{{J}\rho{s}}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{R}}\right)=\frac{\mathrm{3}{T}}{{J}\rho}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{R}}\right)\:\:\:\:\left[\:{s}=\mathrm{1}\right] \\ $$$$ \\ $$

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

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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

most welcome...

$${most}\:{welcome}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com