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 surroindings is

Options

A. a)`10^@C`
B. b)`5^@C`
C. c)`15^@C`
D. d)`20^@C`

Answer

A

Explanation

According to Newton.s law of cooling
`(theta_1- theta_2)/t=k[(theta_1 +theta_2)/2- theta_0]`
In the first case
`implies (60-50)/10=K[(60+50)/2- theta_0]`
`implies 1=K(55- theta)`
In the second case
`implies (50-42)/10=K[(50+42)/2- theta_0]`
`implies 0.8=K[46- theta]`....(ii)
Dividing Eq. (i) by Eq. (ii) ,we get
`1/0.8=(55- theta)/(46- theta)`
or `46- theta=44-0.8 theta`
`theta=10^@C`