1.

A body cools from `80^(@)C` to `70^(@)C` in 5 minutes and further to `60^(@)C` in 11 minutes. What will be its temperature after 15 minutes from the start ? Also determine the temperature of the surroundings.

Answer» Correct Answer - `[54.4^(@)C; 15^(@)C]`
In first case Change in temperature,
`dT=80-70=10^(@)C`
Time interval, `dt= 5` minutes
Average temperature, `T=(80+70)/(2)=75^(@)C`
Temperature of the surrounds `=T_(0)` (say)
As `(dT)/(dt)= -K(T-T_(0))`
`:. (10)/(5)= -K(75-T_(0))` ....(ii)
In second case Change in temperature,
`dT=70-60=10^(@)C`
Time interval, `dt=11-5=6 min`.
Average temperature, `T=(70+60)/(2)=65^(@)C`
Using, `(dT)/(dt)= -K[T-T_(0)]`
`(10)/(6)= -K[65-T_(0)]` ...(ii)
Dividing (i) by (ii), we have
`(6)/(5)=(75-T_(0))/(65-T_(0))`
On solving, `T_(0)=15^(@)C`
From (i), `(10)/(5)= -K[75-15]= -Kxx60` .....(iii)
In third case Let `T_(1)` be tghe temperature of the body after 15 minutes from start. Change in temperature, `dT=60-T_(1)^(@)C`
Time interval, `dt=15-11 =4 min`
Average temperature, `T=((60+T_(1))/(2)) .^(@)C`
As, ` (dT)/(dt)= -K[T-T_(0)]`
`:. (60-T_(1))/(4)= -K[(60+T_(1))/(2)-15]= -K[(T_(1)+30)/(2)]` ...(iv)
Dividing (iv) by (iii), we have
`[(60-T_(1))/(4)xx(5)/(10)]=((T_(1)+30)/(2))xx(1)/(60)`
`60-T_(1)=(T_(1)+30)/(15)` or `60xx15-15 T_(1)=T_(1)+30`
`900-30= 16 T_(1)` or `T_(1)= 870//16=54.4^(@)C`


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