First method : In first case.
`T_(1) = 80^(@)C, T_(2) = 64^(@)C, t = 5 min`
Let `T_(0)` be the temperature of the surroundings. Using the relation
`2.3026 log_(10) ((T_(1)-T_(0))/(T_(2)-T_(0))) = Kt`
`2.3026 log_(10) = (80-T_(0))/(64-T_(0)) = K xx 5` ..(i)
2nd case, `T_(1) = 64^(@)C, T_(2)= 52^(@)C`,
`t=10- 5 = 5 mi n`
the temperature of surrounding remains unchanged, i,e`T_(0)`
`2.3026log_(10)((64-T_(0))/(52-T_(0))) = Kxx5` ..(ii)
From (i) and (ii) , we have
`2.3026 log_(10)((64-T_(0))/(52-T_(0))) = 2.3026 log (( 80-T_(0))/(64-T_(0)))`
or `log_(10)((64-T_(0))/(52-T_(0)))=log_(10) ((80-T_(0))/(64-T_(0)))`
Taking antilogs, we have `(64-T_(0))/(52-T_(0))=(80-T_(0))/(64-T_(0))`
On solving , `T_(0) = 16^(@)C`
Second method : In first case,
Change in temperature, `dT=80 - 64 = 16^(@)C`
time interval , `dt = 5 min`
Average temperature, `T = (80+64)/(2) = 72^(@)C`
Temperature of surronding , `T_(0)=?`
As, `(dT)/(dt) = -K(T-T_(0))`
`:. 16/5 = -K(72-T_(0))` ...(i)
In second case, change in temperature,
`dT = 64-52 = 12^(@)C`
time temperature, `T=(64+52)/(2)=58^(@)C`
Using ,`(dT)/(dt) = -K(T-T_(0))`, we have
`12/5 = -K(58-T_(0))`
Dividing (i) by (ii), we have
`16/5 xx 5/12 = (72-T_(0))/(58-T_(0))`
on solving, `T_(0) = 16^(@)C`.
