A large number of water droplets each of radius ' $t$ ' combine to form a large drop of Radius ' $R$ '. If the surface tension of water is ' $T$ ' \& mechanical equivalent of heat is ' $\mathrm{J}$ ' then the rise in temperature due to this is

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Aicharya · Accepted Answer

Radius of each droplet $=\mathrm{r}$ Radius of the drop $=\mathrm{R}$ As volume remains constant, $\begin{aligned} & \mathrm{n} 	imes \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3 \ & 	herefore \quad \mathrm{n}=\frac{\mathrm{R}^3}{\mathrm{r}^3} \end{aligned}$ Decrease in surface area $=4 \pi r^2 n-4 \pi R^2 n$ $\begin{aligned} \Delta \mathrm{A} & =4 \pi\left[\mathrm{nr}^2-\mathrm{R}^2ight] \ & =4 \pi\left[\frac{\mathrm{R}^3}{\mathrm{r}^3} \mathrm{r}^2-\mathrm{R}^2ight] \ & =4 \pi \mathrm{R}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}ight] \end{aligned}$ Enery released $\mathrm{W}=\mathrm{T} 	imes \Delta \mathrm{T}$ $\begin{aligned} & 	ext { Heat produced } \mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}} \ & 	ext { H } \end{aligned}$ Heat produced $\mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}$ $\mathrm{Q}=\mathrm{m} \cdot \mathrm{s} \cdot \Delta 	heta$ Put (i) and (iii) into (ii) $\begin{aligned} & \mathrm{m} \cdot \mathrm{S} \Delta 	heta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}ight] \ & \frac{4}{3} \pi \mathrm{R} ho_{	ext {water }} \mathrm{S}_{	ext {water }} \Delta 	heta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}ight] \ & \Delta \mathrm{Q}=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}ight] \end{aligned}$ ~

A large number of water droplets each of radius ' $t$ ' combine to form a large drop of Radius ' $R$ '. If…

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