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Question Number 34759 by Tinkutara last updated on 10/May/18

	Answered by ajfour last updated on 10/May/18

	$\Rightarrow 4 π ρ {L r}^{2} d r ⩽ 8 π S r d r$ $o r r ⩾ \frac{2 S}{ρ L} (a s d r < 0) .$
	Commented by Tinkutara last updated on 11/May/18

	$W h a t i s w r o n g i f I d o$ $\frac{4}{3} π R^{3} ρ L = S \times 4 π r^{2}$ $r = \frac{3 S}{ρ L}$ $?$
	Commented by ajfour last updated on 11/May/18

	$p r o c e s s i s c o n t i n u o u s$ $l i t t l e e n e r g y r e q u i r e d i n a$ $d i f f e r e n t i a l t i m e i n t e r v a l d t$ $f o r e v a p o r a t i o n o f a m a s s d m i s$ $ρ L (4 π r^{2} ∣ d r ∣) = l o s s i n d i f f e r e n t i a l$ $s u r f a c e e n e r g y = S (8 π r ∣ d r ∣)$ $(t h e e v a p o r a t i o n s h o u l d t a k e$ $p l a c e i n e a c h s u c h t i m e i n t e r v a l s)$ $a n d t b i s o v e r a l l e q . o f y o u r s$ $e n t i r e s u r f a c e e n e r g y = e n t i r e$ $e n e r g y r e q u i r e m e n t f o r e v a p o r a t i o n$ ${i s n}^{'} t c o r r e c t .$
	Commented by Tinkutara last updated on 11/May/18
	Can't we equate just initial and final conditions when the liquid drop evaporates completely?
	Commented by ajfour last updated on 11/May/18

	$n a .$
	Commented by Tinkutara last updated on 11/May/18
	Thank you very much Sir! I got the answer. ��
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