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. 4 minutes
B. 2 minutes
C. 1 minute
D. 3 minutes

Answer

A

Explanation

According to Newton's law of cooling ,
`(T_1-T_2)/t=K((T_1+T_2)/2-T_s)`
where `T_s` is the temperature of surroundings.
For the first case,
`T_1=65.5^@C , T_2=62.5^@C , T_s=22.5^@C , t = 1 min `
`:.(65.5^@C-62.5^@C)/(1min)=K((65.5^@C+62.5^@C)/2-22.5^@C)`
`(3^@C)/(1min)=K(41.5^@C) " "...(i)`
For the second case,
`T_1=46.5^@C, T_2=40.5^@C, T_s=22.5^@C, t = ? `
`:.(46.5^@C-40.5^@C)/(1min)=K((46.5^@C+40.5^@C)/2-22.5^@C)`
`(6^@C)/t=K(21^@C)" "...(ii)`
Dividing eqn. (i) by eqn. (ii) , we get
`((3^@C)/(1min))/((6^@C)/t)=(K(41.5^@C))/(K(21^@C))" or "t/(2min)=1.98`
or `t = 3.96 min ~~ 4 min`