How coffe is a mug cools form `90^(@)C to 70^(@)C` in 4.8 minutes. The room temeprature is `20^(@)C.` Applying Newton's law of cooling the time needed to cool I further by `10^(@)C ` should be nealry-

Options

A. `4.2` min
B. `3.8` min
C. `3.2` min
D. `2.4` min

Answer

C

Explanation

Average Newton cooling method
`(70-90)/(4.8)=K[(70+90)/(2)-20]" "....(1)`
`rArr-(20)/(4.8)=K[80-20]rArr-(20)/(4.8)=Kxx60`
for next cooling by `10^(@)` C
`(60-70)/(t)=K[(60+70)/(2)-20] " " ....(2)`
`rArr-(10)/(t)=K[65-20]`
`rArr-(10)/(t)=Kxx45`
`rArr-(10)/(t)=45xx((-20)/(4.8xx60))`
`t=(+10xx3xx4.8)/(45)`
`t=3.2` minutes