Saved Bookmarks
| 1. |
The temperature of an isolated black body falls from `T_(1)` to `T_(2)` in time `t`. Let `c` be a constantA. `t = c [(1)/(T_(2)) - (1)/(T_(1))]`B. `t = c [(1)/(T_(2)^(3)) - (1)/(T_(1)^(2))]`C. `t = c [(1)/(T_(2)^(3)) - (1)/(T_(1)^(3))]`D. `t = c [(1)/(T_(2)^(4)) - (1)/(T_(1)^(4))]` |
|
Answer» Correct Answer - C (3) `ms (dT)/(dt) = - sigam A (T^(4) - 0)` `int_(T_(1))^(T_(2)) (dT)/(T^(4)) = - sigma A int_(0)^(t) dt` `[(1)/(T_(2)^(3)) - (1)/(T_(1)^(3))] = (3 sigma A) t` `t - (1)/((3 sigma A)) [(1)/(T_(2)^(3)) - (1)/(R_(1)^(3))]` `t = c [(1)/(T_(2)^(3)) - (1)/(T_(1)^(3))], c = (1)/(3 sigma A)` |
|