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A hot water cools from 92^(@)C to 84^(@)C in 3 minutes when the room temperature is 27^(@)C. How long will it take for it to cool from 65^(@)C to 60^(@)C?

Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/hot-1029680" style="font-weight:bold;" target="_blank" title="Click to know more about HOT">HOT</a> water cools `<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>^(@)C` in 3 minutes. The average temperature of `92^(@)C` and `84^(@)C` is `<a href="https://interviewquestions.tuteehub.com/tag/88-339824" style="font-weight:bold;" target="_blank" title="Click to know more about 88">88</a>^(@)C`. This average temperature is `61^(@)C` above room temperature. By using equation, <br/> `(dT)/(T-T_(s))=-(a)/(<a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</a>)dt or (dT)/(dt)=-(a)/(ms)(T-T_(s))`<br/> `(8^(@)C)/("3 min")=-(a)/(ms)(61^(@)C)"...(1)"` <br/> Similarly the average temperature of `<a href="https://interviewquestions.tuteehub.com/tag/65-331005" style="font-weight:bold;" target="_blank" title="Click to know more about 65">65</a>^(@)C and 60^(@)C`. The average temperature is `35.5^(@)C` above the room temperature. Then we can write <br/> `(5^(@)C)/(dt)=-(a)/(ms)(35.5^(@)C)"....(2)"` <br/> By dividing both the equation, we get <br/> `((8^(@)C)/("3 min"))/((5^(@)C)/(dt))=-((a)/(ms)(61^(@)C))/(-(a)/(ms)(35.5^(@)C))` <br/> `(8xxdt)/(3xx5)=(61)/(35.5)` <br/> `dt=(61xx15)/(35.5xx8)=(915)/(284)="3.22 min"`</body></html>


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