A cup of coffee cools from $90^{\circ}$C to $80^{\circ}$C in t minutes when the room temperature is $20^{\circ}$C. The time taken by a similar cup of coffee to cool from $80^{\circ}$C to $60^{\circ}$C at the same room temperature $20^{\circ}$C is:
(1) $\frac {13}{10}$t
(2) $\frac {13}{5}$t
(3) $\frac {10}{13}$t
(4) $\frac {5}{13}$t
Solution
Using law of cooling, $ - \frac{{\Delta C}}{{\Delta t}} = k({T_{avg.}} - {T_{Room}})$ we have,
Case I
$ - \frac{{(80 - 90)}}{t} = k\left( {\frac{{90 + 80}}{2} - 20} \right)$
$ \Rightarrow \frac{{10}}{t} = 65k$
$\therefore \frac{2}{t} = 13k$ ..........(*)
Case II
$ - \frac{{(60 - 80)}}{{t'}} = k\left( {\frac{{80 + 60}}{2} - 20} \right)$
$ \Rightarrow \frac{{20}}{{t'}} = 50k$
$\therefore \frac{2}{{t'}} = 5k$ ..........(#)
From (*) & (#),
$\frac{{t'}}{t} = \frac{{13}}{5}$
$\therefore t' = \frac{{13}}{5}t$
Answer: (2)
