Saved Bookmarks
| 1. |
The time taken by a liquid to cool from 65^(@)C to 55^(@)C is 5 minutes and cools to 47^(@)C in the next 5 minutes, calculate the room temperature and the temperature of the body after another 5 minutes. |
|
Answer» Solution :`log((theta_(2)-theta_(0))/(theta_(1)-theta_(0)))=(-K)/(mc)t=-K.T` where `K.=K/(mc)` is a constant `theta_(0)=65^(@)C,theta_(0)=55^(@)C,t=5 "minutes", theta_(0)=?` `"log"((55-theta_(0))/(65-theta_(0)))=K.xx5` …(i) For second part `theta_(1)=55^(@)C, theta_(2)=47^(@)Candt=5"minutes"` `"log"((47-theta_(0))/(55-theta_(0)))=-K.xx5` From the EQUATION (i) and (II) `"log"((55-theta_(0))/(65-theta_(0)))="log"((47-theta_(0))/(55-theta_(0)))` `(55-theta_(0))/(65-theta_(0))=(47-theta_(0))/(55-theta_(0))` `(55-theta_(0))^(2)=(65-theta_(0))(47-theta_(0))` `55^(2)-2xx55xxtheta_(0)+theta_(0)^(2)=65xx47-65theta_(0)-47theta_(0)+theta_(0)^(2)` `3025-110theta_(0)=3055-112theta_(0)` `theta_(0)=15^(@)C` LET `theta` be the temperature after another 5 minutes `"log"(theta-15)/(47-15)="log"((47-15)/(55-15))` `(theta-15)/32=32/(40)` `theta=40.6^(@)C` ANOTHER METHOD From Newton.s law of cooling Rate of cooling `prop`{MEAN difference of temperature between the body and the surroundings `(65-55)/(5) prop((54+55)/2)-theta_(0)` `2 prop 60-theta_(0)` ...(i) `(55-47)/5prop((55+47)/2)-theta_(0)` ...(ii) `8/(5)prop51-theta_(0)` dividing equation (i) and (ii) `(2xx5)/8=(60-theta_(0))/(51-theta_(0))` `510-10theta_(0)=480-8theta_(0)` `theta_(0)=30/(2)=15^(@)C` let `theta` be the temperature after another 5 minutes `(47-theta)/5prop((47+theta)/2)-15` ...(iii) from equation (i)`2prop60-theta_(0)` `2prop60-15` `2prop45` ...(iv) Dividing equation (iii) and (iv) `(47-theta )/(5xx2)=((47+theta)/2-15)/45` `47-theta/10=47/(2xx45)+theta/(2xx45)-15/(45)` Solving,`theta=40.6^(@)C` |
|