The initial temperature of a body is `80^(@)C`. If its temperature falls to `64^(@)C` in `5` minute and in `10` minute to `52^(@)`. Find the temperature of surrounding.

Options

A. `16^(@)C`
B. `26^(@)C`
C. `36^(@)C`
D. `40^(@)C`

Answer

A

Explanation

Using Newton's law of cooling
`(Delta theta)/(Deltat) = K[theta_(av)-theta_(0)]`
`((theta_(1)-theta_(2))/(t)) = K [(theta_(1)+theta_(2))/(2)-theta_(0)]`
In the `1^(st)` case, `(80-64)/(5) = K[(80+64)/(2)-theta_(0)]`
or `3.2 = K[72-theta_(0)]` ...(i)
In the `2^(nd)` case, `(64-52)/(5) = K[(64-52)/(2)-theta_(0)]`
or `2.4 = K(58-theta_(0))`..(ii)
From (i) by (ii), we get `(3.2)/(2.4) = (72-theta_(0))/(58-theta_(0))`
or `185.6-3.2 theta_(0) = 172.8-2.4 theta_(0) or theta_(0) = 16^(@)C`.