1000 drops of water, all of same size join together to form a single drop and the energy released raises the temperature of the drop. Given that ? is the surface tension of water, r the radius of each small drop, p the density of liquid, / the mechanical equivalent of heat. What is the rise in temperature ?

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$T / Jr$

$10 TJ /$

$100 T / r$

none of the above

answer is D.

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Detailed Solution

$\frac{4}{3} {πR}^{3} = n \times \frac{4}{3} {πr}^{3} R = (n)^{1 / 3} r = 10 r decrease in surface alea, ΔA = n (4 {πr}^{2}) - 4 {πR}^{2} ΔA = 1000 (4 {πr}^{2}) - 4 π (10 r)^{2} = 4 {πr}^{2} [1000 - 100] = 900 \times 4 {πr}^{2} Energy released = ΔA \times T = 900 \times 4 {πr}^{2} T / J calorie Let △ θ be the rise in temperature, then \{1000 \times \frac{4}{3} {πr}^{3}\} \times 1 \times △ θ = 900 \times 4 {πr}^{2} T / J △ θ = [2 \cdot 7 T / Jr]$

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