A body takes 4 minutes to cool from `100^(@)C` to `70^(@)C`. To cool from `70^(@)C` to `40^(@)C` it will take (room temperture os `15^(@)C`)

Options

A. 14 s
B. 8 s
C. 10 s
D. 5 s

Answer

A

Explanation

According to Newton's law of cooling law,
`(theta_(1)-theta_(2))/(t)=K((theta_(1)+theta_(2))/(2)-theta_(0))` where, `theta_(0)` = temperature of surrounding .
Case I
`(100-70)/(8)=K((100+70)/(2)-15)`
`"or " (15)/(4)=K`(70) ...(i)
Case II
`(70-40)/(t)=K((70+40)/(2)-15)`
`"or " (30)/(t)=K(40)` ...(ii)
Dividing Eq .(i)by Eq.(ii) , we have
`(15)/(4)xx(t)/(30)=(70)/(40)rArrt=(7)/(4)xx(4xx30)/(15)`
`therefore" " t=14 s`