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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» <html><body><p></p>Solution :Radius of larger drop =R<br/>Radius of smaller drop =r<br/>Volume of large drop =Volume of small drops N<br/>`(4)/(3)piR^(3)=N((4)/(3)pir^(3))`<br/>`thereforeR^(3)=Nr^(3)`<br/>`N=(R^(3))/(r^(3))`….(1)<br/>Change in area ,<br/>`DeltaA` = area of large drop - area of small drops N<br/>`=4piR^(2)-N(4pir^(2))`<br/>`=4Deltapi(R^(2)-Nr^(2))`<br/>`therefore` Energy relased `=SDeltaA`<br/>`=S4pi(R^(2)-Nr^(2))`....(2)<br/>(Where S =<a href="https://interviewquestions.tuteehub.com/tag/surface-1235573" style="font-weight:bold;" target="_blank" title="Click to know more about SURFACE">SURFACE</a> tension)<br/>Due to this energy relased ,temperature decrease in temperature is `DeltaQ`.<br/>Then energy <a href="https://interviewquestions.tuteehub.com/tag/released-613817" style="font-weight:bold;" target="_blank" title="Click to know more about RELEASED">RELEASED</a>,<br/>`Q=CmDeltatheta`<br/>`=C((4)/(3)piR^(3)rho)Deltatheta` ...(3)<br/>(Where m = volume V `xx` density `rho=(4)/(3)piR^(3)rho)`<br/>Equating equation (1) and (2),`Sxx4pi(R^(2)-Nr^(2))=((4)/(3)piR^(3)rho)CxxDeltatheta`<br/>`thereforeDeltatheta=(Sxx4pi(R^(2)-Nr^(2)))/(((4)/(3)piR^(3)rho)C)`<br/>`Deltatheta=(<a href="https://interviewquestions.tuteehub.com/tag/3s-311333" style="font-weight:bold;" target="_blank" title="Click to know more about 3S">3S</a>)/(rhoC)[(R^(2)-Nr^(2))/(R^(3))]`<br/>`Deltatheta=(3S)/(rhoC)[(1)/(R)-(Nr^(2))/(R^(3))]`<br/><a href="https://interviewquestions.tuteehub.com/tag/putting-2963476" style="font-weight:bold;" target="_blank" title="Click to know more about PUTTING">PUTTING</a> the <a href="https://interviewquestions.tuteehub.com/tag/value-1442530" style="font-weight:bold;" target="_blank" title="Click to know more about VALUE">VALUE</a> of N from equation (1),<br/>`Deltatheta=(3S)/(rhoC)[(1)/(R)-(R^(3))/(r^(3))((r^(2))/(R^(3)))]`<br/>`=(3S)/(5C)[(1)/(R)-(1)/(r)]`....(4)<br/>decrease in temperature.</body></html>


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