1.

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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