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A body cools in 10 min from 60^(@)C to 40^(@)C in 5 min and to 42.5^(@)C in 7.5 min. Find the temperature of the surroundings.

Answer» <html><body><p></p>Solution :According to Newton's <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> of cooling. <br/> `((T_(1)-T_(2))/(t))= <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>[((T_(1)+T_(2))/2)-T_(0)]` <br/> For the given conditions, <br/> `(60-40)/(10) = k[(60-40)/(2)-10]`.....................(i) <br/> Let T be the temperature after next 10 min. Then, <br/> `(40-T)/(10)=K[(40+T)/(2)-10]`..............(ii) <br/> <a href="https://interviewquestions.tuteehub.com/tag/solving-1217196" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVING">SOLVING</a> Eqs. (i) and (ii), we get T=`<a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a>^(@)`C</body></html>


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