Physics - Fluid Question with Solution | TestHub

Radius of each droplet $=\mathrm{r}$ Radius of the drop $=\mathrm{R}$ As volume remains constant, $$\begin{array}{rl}& \mathrm{n}\times \frac{4}{3}\pi {\mathrm{r}}^3=\frac{4}{3}\pi {\mathrm{R}}^3\\ & \therefore \mathrm{n}=\frac{{\mathrm{R}}^3}{{\mathrm{r}}^3}\end{array}$$ Decrease in surface area $=4\pi r^{2}n-4\pi R^{2}n$ $$\begin{array}{rl}\Delta \mathrm{A}& =4\pi \left[{\mathrm{nr}}^2-{\mathrm{R}}^2\right]\\ & =4\pi \left[\frac{{\mathrm{R}}^3}{{\mathrm{r}}^3}{\mathrm{r}}^2-{\mathrm{R}}^2\right]\\ & =4\pi {\mathrm{R}}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]\end{array}$$ Enery released $\mathrm{W}=\mathrm{T}\times \Delta \mathrm{T}$ $$\begin{array}{rl}& \text{ Heat produced }\mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}\\ & \text{ H }\end{array}$$ Heat produced $\mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}$ $$\mathrm{Q}=\mathrm{m}\cdot \mathrm{s}\cdot \Delta \theta$$ Put (i) and (iii) into (ii) $$\begin{array}{rl}& \mathrm{m}\cdot \mathrm{S}\Delta \theta =\frac{4\pi {\mathrm{R}}^3\mathrm{T}}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]\\ & \frac{4}{3}\pi \mathrm{R}{\rho }_{\text{water }}{\mathrm{S}}_{\text{water }}\Delta \theta =\frac{4\pi {\mathrm{R}}^3\mathrm{T}}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]\\ & \Delta \mathrm{Q}=\frac{3\mathrm{T}}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]\end{array}$$