A copper ball cools from 62^(@)C to 50^(@)C in 10 minutes and to 42^(@)C in the next 10 minutes. Calculate its temperature at the end of another 10 minutes.

A copper ball cools from 62^(@)C to 50^(@)C in 10 minutes and to 42^(@

1.	A copper ball cools from 62^(@)C to 50^(@)C in 10 minutes and to 42^(@)C in the next 10 minutes. Calculate its temperature at the end of another 10 minutes.
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`(theta_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)-theta_(2))/t_(1)=<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>[(theta_(1)+theta_(2))/2-theta_(0)]` <br/> `(62-50)/10=k[(62-50)/2-theta_(0)]" "...(1)` <br/> `(theta_(2)-theta_(3))/t_(2)=k[(theta_(2)+theta_(3))/t_(2)-theta_(0)]` <br/> `(50-42)/10=k[(50+42)/(2)-theta_(0)]""...(2)` <br/> `((1))/((2))" gives "theta_(0)=<a href="https://interviewquestions.tuteehub.com/tag/26-298265" style="font-weight:bold;" target="_blank" title="Click to know more about 26">26</a>^(@)C` <br/> `(theta_(3)-theta_(4))/t_(3)=k[(theta_(3)+theta_(4))/2-theta_(0)]` <br/> `(42-theta_(4))/10=k[(42-theta_(4))/2-26]rArrtheta_(4)=36.7^(@)C`</body></html>