Hot water cools from `60^@C` to `50^@C` in the first 10 min and to `42^@C` in the next 10 min. The temperature of the surrounding is

Options

A. `20^(@)` C
B. `30^(@)` C
C. `15^(@)` C
D. `10^(@)` C

Answer

D

Explanation

According to Newton.s law of cooling
`(theta_(2) - theta_(1))/t = K[(theta_(1) + theta_(2))/2 - theta_(s)]`
Where, `theta_(s)` is the temperature of the surroundings.
`(60 - 50)/10 = K[(60 + 50)/2 - theta_(s)]`
`1 = K[55- theta_(s)]`........ (i)
Similarly, `(50 - 42)/10 = K(46 - theta_(s))`
`8/10 = K(46 - theta_(s))`......... (ii)
Dividing Eq. (i) by Eq. (ii), we get:
`10/8 = (K(55 - theta_(s)))/(K(46 - theta_(s)))`
`rArr theta_(s) = 10^(@)` C