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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. |
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Answer» SOLUTION :`(theta_(1)-theta_(2))/t_(1)=K[(theta_(1)+theta_(2))/2-theta_(0)]` `(62-50)/10=k[(62-50)/2-theta_(0)]" "...(1)` `(theta_(2)-theta_(3))/t_(2)=k[(theta_(2)+theta_(3))/t_(2)-theta_(0)]` `(50-42)/10=k[(50+42)/(2)-theta_(0)]""...(2)` `((1))/((2))" gives "theta_(0)=26^(@)C` `(theta_(3)-theta_(4))/t_(3)=k[(theta_(3)+theta_(4))/2-theta_(0)]` `(42-theta_(4))/10=k[(42-theta_(4))/2-26]rArrtheta_(4)=36.7^(@)C` |
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