A pan filled with hot food cools from `94^(@)C` to `86^(@)C` in 2 minutes when the room temperature is at `20^(@)C`. How long will it take to cool from `71^(@)C` to ` 69^(@)C`? Here cooling takes place according to Newton's law of cooling.

Options

A. 50 s
B. 40 s
C. 38 s
D. 42 s

Answer

D

Explanation

d) From the Newton's law of cooling,
`(theta_(1)-theta_(2))/t alpha((theta_(1)+theta_(2))/2-theta_(0))`
Where, `theta_(0)` = temperature of surrounding
`therefore` `(94-86)/(t_(1))alpha((theta_(1)+theta_(2))/2-theta_(0))`
Where, `theta_(0) = temperature of surrounding
`therefore` `therefore`(94-86)/t_(1)alpha((94+86)/2-20)`
`8/t_(1)alpha(70)`...............(i)
`(71-69)/(t_(2))alpha((71+69)/2-20)`
`2/(t_(2))alpha50`..............(ii)
Dividing Eq. (ii) by Eq. (i), we get
`t_(2)/t_(1) = (70)/(50xx4) rArr t_(2) = (70xx2)/(50xx4)`
`t_(2)` = 0.7 min `rArr t_(2)` = 42s