question_answer 50)

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

Answer:

                  Volume of drop of radius \[R=N\times \]volume of drop of radius \[r\]                 \[\frac{4}{3}\,\pi {{R}^{3}}=\,N\times \,\frac{4}{3}\,\pi \,{{r}^{3}}\] or \[r=\,\frac{R}{{{(N)}^{1/4}}}\]                 Change is surface area, \[\Delta A\,=\,4\pi \,({{R}^{2}}-N{{r}^{2}})\]                 \[=\,4\pi \,({{R}^{2}}-\,{{N}^{\frac{1}{3}}}{{R}^{2}})\,\,=\,4\pi \,{{R}^{2}}\,(1-\,{{N}^{1/3}})\]                 \[\therefore \] Energy utilised \[=4\pi \,{{R}^{2}}\,(1-{{N}^{1/3}})\,T\,\]                 Let \[\Delta \theta \] be the decrease in temperature therefore, energy utilised \[=\,m\,C\Delta \theta \],                 where C is the specific heat of liquid.                 Now \[mc\,\,\Delta \theta =\,4\pi \,{{R}^{2}}\,(1-{{N}^{1/3}})\,T\]                 or            \[\Delta \theta =\,\frac{4\pi \,{{R}^{2}}\,(1-\,{{N}^{1/3}})T}{mc}\]                 Since mass of drop, \[m=\,\frac{4}{3}\,\pi \,{{R}^{3}}\rho \]                 \[\therefore \] \[\Delta \theta =\,\frac{3\,\times \,4\,\pi \,{{R}^{2}}(1-\,{{N}^{1/3}})T}{4\pi \,{{R}^{3}}\,\rho C}\]                 \[=\,\frac{3T}{R\rho C}\,(1-\,{{N}^{1/3}})\].