Saved Bookmarks
| 1. |
A solid copper sphere (density `=8900kgm^(-3)` and specific heat `C=390Jkg^(-1)K^(-1)`) of radius `r=10cm` is at an initial temperature `T_(1)=200K`. It is then suspended inside an chamber whose walls are at almost `OK`. Calculate the time required for the temperature of the sphere to drop to `T_(2)=100 K`, `sigma=5.67xx10^(-8)Wm^(-2)K^(-4)`. |
|
Answer» Let `T` be the temperture at any time `t`. Then rate of loss of heat `=((4pi)/(3)r^(3)rho)c (-(dT)/dt)`. Also the rate of loss of heat `=sxx4pir^(2)(T^(2)-0)` `:.-(4pi)/(3)r^(3)rhoc(dT)/(dt)=4pr^(2)sT^(4)impliesdt=(rhorc)/(3sigma)(dT)/(dT^(4))` `implies t=(rhorc)/(3sigma)int_(T_(1))^(T_(2))(dT)/(T^(4))=(rhorc)/(9sigma)(T_(1)^(3)-T_(2)^(3))/(T_(1)^(3)T_(2)^(3))` `impliest=(10xx10^(-2)xx8900xx390)/(9xx5.67xx10^(-8))xx(200^(3)-100^(3))/(200^(3)xx100^(3))=595165s=165h19min` |
|