Saved Bookmarks
| 1. |
A block of metal is heated to a temperature much higher then the room temperature and placed in an evacuated cavity. The curve which correctly represents the rate of cooling ( T is temperature of the block and t is the time ) |
|
Answer»
`(dT)/dt=k.(T - T_(0))` where `k. = k/(ms) ` and negative sign shows the rate of heat loss. The given expression can be rearranged by integrating as, `int (dT)/(T-T_(0))= - k.int dt` `log_(e) (T-T_(0))=-k.t + log_(e)A (because log_(e) A =" Constant")` `-(dT)/(T-T_(0))=k.dt` `rArr ln (T-T_(0))=-k.t` `rArr T = e^(-k.t)+T_(0)" at " t rarr infty` `rArr T=T_(0)," at " t = 0 rArr T rarr infty ` Hence, the graph as shown below, shows TEMPERATURE of a body (done) varies exponentially with time from T to `T_(0) ( T_(0) lt T)`.  THUS, the correct option is (b). |
|
