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A metal ball cools from `62^(@)C` to `50^(@)C` in `10min` and to `42^(@)C` the next ten minutes. What will be its temperature at the end of next ten minutes ? |
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Answer» Use `(theta_(1)-theta_(2))/(t)=K((theta_(1)-theta_(2))/(2)theta_(0))` ………`(1)` to get the following equations `(62-50)/(10)=K((62+50)/(2)-theta_(0))` and `(50-42)/(10)=((50+42)/(2)-theta_(0))`……..`(2)` Divide `2` by `1` and solve to get `theta_(0)` `(8)/(12)=(46-theta_(0))/(56-theta_(0))` `implies theta_(0)=26`.......`(3)` Let after the next `10min` the temperture falls to then `(42-theta)/(10)=K((42+theta)/(2)-26)`......`(4)` From `(3)` using value of `theta_(0)=26` we get `(12)/(10)-k(56-26)` Divide `(4)` by `(5)` to get `(42-theta)/(12)=(42+theta-52)/(2(56-26))` `implies theta=36.7^(@)C` |
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