Certain quantity of water cools from to in the first 5 minutes and to in the next 5 minutes. The temperature of the surroundings is [2014]

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Solution

Q.
Certain quantity of water cools from $70 ° C$ to $60 ° C$ in the first 5 minutes and to $54 ° C$ in the next 5 minutes. The temperature of the surroundings is [2014]

1 $45 ° C$

2 $20 ° C$

3 $42 ° C$

4 $10 ° C$

Ans.
(1)

Let $T_{s}$ be the temperature of the surroundings.

According to Newton’s law of cooling

$\frac{T_{1} - T_{2}}{t} = K (\frac{T_{1} + T_{2}}{2} - T_{s})$

For first 5 minutes,

$T_{1} = 70 ° C,$ $T_{2} = 60 ° C,$ $t = 5$ minutes

$∴$ $\frac{70 - 60}{5} = K (\frac{70 + 60}{2} - T_{s}) = K (65 - T_{s})$ ...(i)

For next 5 minutes,

$T_{1} = 60 ° C,$    $T_{2} = 54 ° C,$ $t = 5$ minutes

$∴$ $\frac{60 - 54}{5} = K (\frac{60 + 54}{2} - T_{s})$

$\frac{6}{5} = K (57 - T_{s})$ ...(ii)

Divide eqn. (i) by eqn. (ii), we get

   $\frac{5}{3} = \frac{65 - T_{s}}{57 - T_{s}}$

   $285 - 5 T_{s} = 195 - 3 T_{s}$

   $2 T_{s} = 90 \Rightarrow T_{s} = 45 ° C$

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