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A solid sphere of radius R is made of a material of density rhoand specific heat c. The sphere is initially at 220 K and is later suspended inside a chamber whose walls are at 0K. What time is required for the temperature of the sphere to drop to 160 K?

Answer» <html><body><p></p>Solution :The energy radiated by the sphere is given by Stefan.s law , <br/> ` (dQ)/(dt) = sigma AT^4 = sigma (4pi R^2) T^4` <br/> ` = 4pi sigma R^2 T^4`.....(i) <br/>Due to energy loss, the temperature of the sphere is reduced. Rate of energy loss due to rate of decrease of temperature is <br/> `(dQ)/(dt) = mc (dT)/(dt)` <br/> `m= <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> rho = 4/3 pi R^3 rho ` <br/> ` therefore (dQ)/(dt) =- 4/3 pi R^3rho c ( (dT)/(dt) )`..(ii) <br/>Comparing (i) and (ii) <br/> ` 4pi sigma R^2 T^4 = - 4/3 pi R^3 rho c ((dT)/(dt) )` <br/> `rArr 3 sigma T^4 = - rho <a href="https://interviewquestions.tuteehub.com/tag/rc-613378" style="font-weight:bold;" target="_blank" title="Click to know more about RC">RC</a> ((dT)/(dT))` <br/> ` rArr (rho R c)/(T^4) dT= -3sigma dt` <br/><a href="https://interviewquestions.tuteehub.com/tag/integrating-2131700" style="font-weight:bold;" target="_blank" title="Click to know more about INTEGRATING">INTEGRATING</a> both sides, we get <br/>`rho Rc int_(T_1)^(T_2) T^(-4) dT = - 3 sigma int_(0)^(t) dt` <br/>where `T_1` and `T_2`are initial and final temperature of the sphere, respectively, and t is the time taken for the sphere to drop its temperature from ` T_1 "to" T_2`<br/> ` rArr t = (rho Rc)/(9 sigma) [ (1)/(T_2^3) - (1)/(T_1^3) ]` <br/> ` =(rho Rc)/(9 (5.67 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(-<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)) ) [ (1)/(160^3) - (1)/(220^3) ]` <br/>= 0.3`rhoRc`</body></html>


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