A large number of water drops,each of radius | vedclass

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(C) Let $n$ be the number of small drops of radius $r$ that combine to form a large drop of radius $R$.The volume is conserved: $n \cdot \frac{4}{3} \

Vedclass · Accepted Answer

$\frac{3T}{J} \left( \frac{1}{r} - \frac{1}{R} \right)$

Vedclass · Answer

(C) Let $n$ be the number of small drops of radius $r$ that combine to form a large drop of radius $R$.The volume is conserved: $n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$,so $n = \frac{R^3}{r^3}$.The change in surface area is $\Delta A = n(4 \pi r^2) - 4 \pi R^2 = 4 \pi (n r^2 - R^2)$.Substituting $n = \frac{R^3}{r^3}$,we get $\Delta A = 4 \pi \left( \frac{R^3}{r} - R^2 ight) = 4 \pi R^3 \left( \frac{1}{r} - \frac{1}{R} ight)$.The energy released is $W = T \cdot \Delta A = 4 \pi R^3 T \left( \frac{1}{r} - \frac{1}{R} ight)$.This energy is converted into heat: $Q = mS \Delta 	heta$,where $m = ho \cdot \frac{4}{3} \pi R^3$.Using $W = JQ$,we have $4 \pi R^3 T \left( \frac{1}{r} - \frac{1}{R} ight) = J \cdot ho \cdot \frac{4}{3} \pi R^3 \cdot S \cdot \Delta 	heta$.For water,density $ho = 1 	ext{ g/cm}^3$ and specific heat $S = 1 	ext{ cal/g}^{\circ}	ext{C}$.Thus,$\Delta 	heta = \frac{3T}{J} \left( \frac{1}{r} - \frac{1}{R} ight)$.

$A$ large number of water drops,each of radius $r$,combine to form a single drop of radius $R$. If the surface tension is $T$ and the mechanical equivalent of heat is $J$,then the rise in temperature will be:

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