Certain quantity of water cools from `70^(@)C` to `60^(@)C` in the first 5 min and to `54^(@)C` in the next 5 min. The temperature of the surrounding is

Options

A. `45^(@)`C
B. `20^(@)`C
C. `42^(@)`
D. `10^(@)`C

Answer

A

Explanation

Let temperature of surrounding be `Q_(0)`, then from Newton's law of cooling
Case I = `(Q_(2)-Q_(1))/t = K((Q_(2)+Q_(1))/2-Q_(0))`
`(70-60)/5 = (65-Q_(0))`................(i)
Case II `(60-54)/5 = (57 - Q_(0))K`
From Eqs. (i) and (ii)
`(10/6) = (65-Q_(0))/(57-Q_(0)) rArr Q_(0) = 45^(@)`C