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A liquid cools from `70^(@)C` to `60^(@)C` in `5` minutes. Calculate the time taken by the liquid to cool from `60^(@)C` to `50^(@)C` , If the temperature of the surrounding is constant at `30^(@)C` . |
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Answer» Correct Answer - `7 minutes` `(70 -60)/(5) = K [(70 +60)/(2)-30]` `rArr (10)/(5) = K [65 - 30] ….(i)` Now `(60 -50)/(t) =K [65 - 30] …(ii)` Dividing equation (i) and (ii) `(t)/(5) = (35)/(25)` `t = (7)/(5) xx5 = 7 min` |
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