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If a liquid takes 30 seconds in cooling ...

If a liquid takes `30` seconds in cooling of `80^(@)C` to ` 70^(@)C` and `70` seconds in cooling `60^(@)C` to `50^(@)C`, then find the room temperature.

Text Solution

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`(theta_(1)-theta_(2))/t = K((theta_(1)+theta_(2))/(2) -theta_(0))`
In the first case, `(80-70)/(3)0 = K((80+70)/(2) - theta_(0)) , 1/3= K(75-theta_(0)) …. (i)`
In the second case , ` (60-50)/(70) = K((60+50)/(2) - theta_(0)) , (1)/(7) = K(55-theta_(0)) … (ii)`
Equation `(i)` divide by equation `(ii) 7/3 = ((75-theta_(0)))/((55-theta_(0))) implies 385-7theta_(0) = 225-3theta_(0) implies theta_(0) = 160/4 = 40^(@)C`
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