Hot water cools from `60^(@)C` to `50^(@)C` in the first 10 minutes and to `42^(@)C` in the next 10 minutes Then 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)`
`1=K(55-theta_(s))` . . . (i)
`(50-42)/(10)=K(46-theta_(s))`
`(50-42)/(10)=K(46-theta_(s))`
`8/(10)=K(46-theta_(s))` . . .(ii) Divide (i) by (ii), we get
`(10)/(8)=K(46-theta_(s))` . . .(ii)
Divide (i) by (ii),we get
`(10)/(8)=(K(55-theta_(s)))/(K(46-theta_(s)))`
`10(46-theta_(s))=8(55-theta_(s))`
`460-10theta_(s)=440-8theta_(s)`
`2theta_(s)=20`
`theta_(s)=10^(@)C`