If a drop of liquid breaks into smaller droplets, it result in lowering of temperature of the droplets. Let a drop of radius R, breaks into N small droplets each of radius r. Estimate the drop in temperature.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Volume of a big drop of radius R = N xx volume of small drop let of radius r `:. (4)/(3) pi R^3 = N xx(4)/(3)pi r^3 or R^3 = N r^3 or N = R^3//r^3`
Change in surface area `Delta U = ST xx` change in surface area `= Txx 4pip[R^2 – N r^2]` All this energy relased is at the cost of loweiring the temperature. Mass of the big drop of liquid, `m =(4)/(3) pi R^3 rho` Let s be the specific heat of liquid drop, `Delta theta` the decrease in temperature.
Then `Delta theta = (Delta U)/(ms) = (Txx4pi(R^2 – Nr^2))/((4)/(3) pi R^3 rho)s = (3T)/(rho s) [(1)/(R) – (N r^2)/(R^3)] = (3T)/(rho s)[(1)/(R) – (R^3)/(r^3) xx(r^2)/(R^3)]`
`= (3T)/(rho)[(1)/(R) – (1)/(r)]`