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A copper ball cools from 62^(@)C , C to ...

A copper ball cools from `62^(@)`C , C to `50^(@)`C in 10 minutes and to `42^(@)`C ill the next 1 Ominutes. Calculate its temperature at the end of allother 10 minutes.

Text Solution

Verified by Experts

`(theta_(1) - theta_(2) )/(t_(1)) = k [ (theta_(1) + theta_(2))/(2) - theta_(0) ]`
`(62 - 50)/(10) = K [ (62 + 50)/(2) - theta_(0) ] ` - (i)
`(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 ] rArr theta_(4) = 36.7^(@) C `
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