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(B) When two identical drops of radius $r$ combine to form a single drop of radius $R$,the volume remains constant: $2 	imes (4/3) \pi r^3 = (4/3) \p

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(B) When two identical drops of radius $r$ combine to form a single drop of radius $R$,the volume remains constant: $2 	imes (4/3) \pi r^3 = (4/3) \pi R^3$,which gives $R = 2^{1/3} r$. The initial surface area is $2 	imes (4 \pi r^2) = 8 \pi r^2$. The final surface area is $4 \pi R^2 = 4 \pi (2^{1/3} r)^2 = 4 \pi (2^{2/3}) r^2 \approx 6.35 \pi r^2$. Since the surface area decreases,the total surface energy of the system decreases. According to the law of conservation of energy,this released surface energy is converted into internal energy,which causes the temperature of the drop to increase.

If two identical mercury drops are combined to form a single drop,then its temperature will

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