InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A body initially at `80^(@)C` cools to `64^(@)C` in 5 minutes and to `52^(@)C` in 10 minutes. What is the temperature of the surroundings? |
|
Answer» 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`. |
|