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A hot liquid kept in a beaker cools from `80^(@)C` to `70^(@)C` in two minutes. If the surrounding temperature is `30^(@)C`, find the time of coolilng of the same liquid from `60^(@)C` to `50^(@)C`. |
Answer» Correct Answer - `[216 s]` As, `(dT)/(dt)= -K[(T_(1)+T_(2))/(2)-T_(0)]` `:. (80-70)/(2xx60)= -K[(80+70)/(2)-30]` or `(10)/(2xx60)= -Kxx45` And ` (60-50)/(t)= -K[(60+50)/(2)-30]` or `(10)/(t)= -Kxx25` …(ii) Dividing (i) by (ii), we get `(t)/(2xx60)=(45)/(25)=(9)/(5)` or `t=216 s` |
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