A cup of tea cools from `65.5^@C " to " 62.5^@C` in 1 minute in a room at `22.5^@C` How long will it take to cool from `46.5^@C " to " 40.5^@C` C in the same

Options

A. 1
B. 3
C. 2
D. 4

Answer

D

Explanation

According to Newton.s law of cooling
`(theta_(1)-theta_(2))/(t)=k[(theta_(1)+theta_(2))/(2)-theta_(0)]` where `theta_(0)` is temp. of surroundings.
`:." "(65*5-32*5)/(1)=k[(65*5+62*2)/(2)-22*5]` . . . (i)
Also `(46*5-40*5)/(t)=k[(46*5+40*5)/(2)-22*5]` . . . (ii)
Dividing (i) by (ii)
`t/2=(64-22*5)/(43*5-22*5)`
`rArrt=4` min.
Correct choice is (d).