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A liquid initally at `70^(@)C` cools to `55^(@)C` in 5 minutes and `45^(@)C` in 10 minutes. What is the temperature of the surrounding ? |
Answer» Correct Answer - `[25^(@)C]` 1st case, change in temperature, `dT=70-55=15^(@)C` time interval, dt=5 min Average temperature, `T=(70+55)/(2)=62.5^(@)C` Temperature of surrounding, `T_(0)=?` As, `(dT)/(dt)=-K(T-T_(0))` `:. (15)/(5) = - K(62.5-T_(0))` .....(i) 2nd case, `dT=55-45=10^(@)C`, `dt=10-5=5 min` `T=(55+45)/(2)=50^(@)C` Then, `(10)/(5)= -K(50 - T_(0))` ....(ii) Dividing (i) by (ii) and solving we get, `T_(0)=25^(@)C` |
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