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. `25^@C`
B. `10^@C`
C. `15^@C`
D. `20^@C`

Answer

B

Explanation

From Newton.s law of cooling
`(theta_1 - theta_2)/(t ) = -K [ (theta_1 + theta_2)/(2) - theta_0 ] `
where ` theta_0` is the temperature of surrounding. Initially, hot water cools from `60^@C` to `50^@C` in 10 minutes,
`(60 -50)/(10) = -K [ (60 + 50)/(2) - theta_0 ] ` ....(i)
Again, it cools from `50^@C` to `42^@C` in next 10 minutes.
` (50 - 42)/(10) = -K [ (50 + 42)/(10) - theta_0] ` ....(ii)
Dividing equation (i) by (ii) we get
` (1)/(0.8) = (55- theta_0)/(46 - theta_0)`
`460 - 10 theta_0 = 440 - 8 theta_0`
` 2 theta_0 = 20`
` theta_0 10^@C`