A cup of coffee cools from to in minutes, when the room temperature is . The time taken by a similar cup of coffee to cool from to at a room temperature same at is [2021]

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Solution

Q.
A cup of coffee cools from $90 ° C$ to $80 ° C$ in $t$ minutes, when the room temperature is $20 ° C$ . The time taken by a similar cup of coffee to cool from $80 ° C$ to $60 ° C$ at a room temperature same at $20 ° C$ is [2021]

1 $\frac{5}{13} t$

2 $\frac{13}{10} t$

3 $\frac{13}{5} t$

4 $\frac{10}{13} t$

Ans.
(3)

Initial temperature $(T_{i}) = 90 ° C$

Final temperature $(T_{f}) = 80 ° C$

Room temperature $(T_{0}) = 20 ° C$

Let time taken be $t$ minutes.

According to Newton’s law of cooling,

Rate of cooling $\frac{d T}{d t} = K [\frac{(T_{i} + T_{f})}{2} - T_{0}]$

   $\frac{(T_{f} - T_{i})}{t} = K [\frac{(90 + 80)}{2} - 20] \Rightarrow \frac{90 - 80}{t} = K [65]$

   $K = \frac{10}{65 t}$

In 2nd condition,

Initial temperature $T_{i} = 80 ° C$

Final temperature $T_{f} = 60 ° C$

Let time taken be $t'$ minutes.

Then, $\frac{(80 - 60)}{t'} = \frac{10}{65 t} [\frac{(80 + 60)}{2} - 20]$    $\frac{20}{t'} = \frac{10}{65 t} (50)$

$\Rightarrow t' = \frac{13}{5} t$

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