Saved Bookmarks
| 1. |
The excess temperature of a hot body above its surroundings is halved in t = 10 minutes. In what time will it be (1)/(10) of its initial value. Assume Newton's law of cooling. |
|
Answer» Solution :USING NEWTON's law of cooling we have `(T-T_(s)) = (T_(0) -T_(s))e^(-kt)` at `t =10 min(T-T_(s)) = (T_(0)-T_(s))/(2)` `RARR e^(-k(10)) =(1)/(2)` or `k = (In(2))/(10)` Then at `t = t_(0)` `T-T_(s) =(T_(0)-T_(s))/(10)` `rArr e^(-kt)0 = (1)/(10)` `t_(0) (In(10))/(In(2)) xx 10` =33.23 min |
|