TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

According to Stefan's law heat energy absorbed is given by $\frac{dQ}{dt}=\varepsilon\sigma A(T^{4}-T_{0}^{4})$....(i) where, $T=$ temperature of body and $T_{0}=$ temperature of surroundings. If $m$ be the mass of body and $c$ be the specific heat, then heat loss at temperature $T$ is $\frac{dQ}{dt} = -m c \frac{dT}{dt}$...(ii) From Eqs. (i) and (ii), we get $-m c\;\frac{dT}{dt}=\varepsilon\sigma A(T^{4}-T_{0}^{-4})$ $\Rightarrow\;\;-\;\frac{dT}{dt}=\;\frac{\varepsilon\sigma A}{mc}(T^{4}-T_{0}^{4})$ Here, $T=T_{0}+\Delta T$ $-\;\frac{dT}{dt}=\;\frac{\varepsilon\sigma A}{mc}[(T_{0}+\Delta T)^{4}-T_{0}^{4}]$ $=\;\frac{\varepsilon\sigma A}{mc}T_{0}^{4}\left[\left(1+\frac{\Delta T}{T_{0}}\right)^{4}-1\right]$ $= \frac{\varepsilon\sigma A}{mc} T_{0}^{4}\left[1+\frac{4\Delta T}{T_{0}}-1\right](\because \Delta T\ll T_{0})$ $=\;\frac{\varepsilon\sigma A 4 T_{0}^{3}}{mc}\Delta T$ $\Rightarrow\;\;\;-\;\frac{dT}{dt}=k\Delta T$ wherc, $k=\;\frac{\varepsilon\sigma A 4 T_{0}^{3}}{mc}$ $\therefore\;\; \;\frac{dT}{dt}=-k(T-T_{0})$ This is Newton's law of cooling. Hence, Newton's law of cooling is a special case of Stefan's law.