A liquid cools from `50^(@)C` to `45^(@)C` in 5 minutes and from `45^(@)C` to `41.5^(@)C` in the next 5 minutes. The temperature of the surrounding is

Options

A. `27^(@)C`
B. `40.3^(@)C`
C. `23.3^(@)C`
D. `33.3^(@)C`

Answer

D

Explanation

Applying `(theta_(1)-theta_(2))/t = alpha[(theta_(1)+theta_(2))/2 - (theta_(0))]` two times, we get where alpha is constant.
`(50-45)/5 = alpha[(50+45)/2-(theta_(0))`...............(i)
`(45-41.5)/(5) = alpha[(45+41.5)/2-(theta_(@))]`.......................(ii)
Solving these two equations, we get
`theta_(@)`= temperature of atmosphere = `33.3^(@)`C