A body cools from `60^(@)C` to `50^(@)C` in 10 minutes when kept in air at `30^(@)C` . In the next 10 minut es its temperature will be

Options

A. Below `40^(@)C`
B. `40^(@)C`
C. Above `40^(@)C`
D. Cannot be predicted

Answer

C

Explanation

`(60-50)/(10) = K [(60 + 50)/(2) - 30] "….."(1)`
`(50 - theta)/(10) = K[(50+theta)/(2) - 30] "…."(2)`
By `(2)//(1)`
`(50-theta)/(10) = (theta - 10)/(50) = theta = 43.3^(@)C`