Ncert Exercises & Solutions

Q. 19 If a drop of liquid breaks into smaller droplets, it results in a lowering of the temperature of the droplets. Let a drop of radius R breaks into N small droplets each of radius r. Estimate the drop in temperature.

Hint: The releasing of energy results in the lowering of the temperature.

Step 1: Find the change in surface area.

When a big drop of radius R breaks into N droplets each of radius r, the volume remains constant.

∴

$∴$ The volume of big drop = N x volume of each small drop

\frac{4}{3} {πR}^{3} = N \times \frac{4}{3} {πr}^{3}

$\frac{4}{3} {πR}^{3} = N \times \frac{4}{3} {πr}^{3}$

R^{3} = N r^{3}

$R^{3} = N r^{3}$

N = \frac{R^{3}}{r^{3}}

$N = \frac{R^{3}}{r^{3}}$

Now, change in surface area,

$∆ A = 4 {πR}^{2} - N \times 4 {πr}^{2} = 4 π (R^{2} - {Nr}^{2})$

Step 2: Find the energy released.

The energy released=

$T \times ∆ A = T \times 4 π (R^{2} - {Nr}^{2})$ [T=Surface tension]

Step 3: Find the change in temperature.

$ρ$ is the density and s is the specific heat of liquid and its temperature is lowered by

$∆ θ$ , then the energy released = ms

$∆ θ$

$T \times 4 π (R^{2} - {Nr}^{2}) = (\frac{4}{3} {πR}^{3} \times ρ) s ∆ θ [∴ m = Vρ = \frac{4}{3} {ρπR}^{3}] ∆ θ = \frac{T \times 4 π (R^{2} - {Nr}^{2})}{\frac{4}{3} {πR}^{3} ρ \times s} = \frac{3 T}{ρs} [\frac{R^{2}}{R^{3}} - \frac{{Nr}^{2}}{R^{3}}] = \frac{3 T}{ρs} [\frac{1}{R} - \frac{(R^{3} / r^{3}) \times r^{2}}{R^{3}}] = \frac{3 T}{ρs} [\frac{1}{R} - \frac{1}{r}]$

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