Answer to Question #319027 in Molecular Physics | Thermodynamics for Precious

Question

A copper cube with sides of length 40mm had its surface blackened.the cube is heated up to 500k and place inside a room at a constant temperature of 300k.calculate the initial rate of temperature fall of the cube, assuming it is a perfect blackbody.take SHC of aluminium to be 400j/kg/k, density of aluminium=7800kg/m3

Answer 1

$\frac Q {\Delta t S}=\frac{c\rho V\Delta T}{\Delta tS}=\sigma (T_1^4-T_2^4),$

$\frac{\Delta T}{\Delta t}=\frac{6\sigma (T_1^4-T_2^4)}{c\rho a}=0 .15~\frac Ks.$