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A cup of coffee cools from $90^{\circ}$C ....

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)

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