A cubical block of mass 1.0 kg and edge 5.0 cm is heated to 227^ C . It is placed in an evacuated chamber maintained at 27^ C . Assuming the block emits radiation as a blackbody, determine the rate at which the block's temperature will decrease. The specific heat capacity of the block's material is 400 J kg ⁻¹ K ⁻¹ .

Step-by-step solution

The surface area of the cubical block is A = 6a^2 . Substituting a = 0.05 m , we get A = 6 (0.05)^2 = 0.015 m ^2 . According to the Stefan-Boltzmann law, the net rate of heat loss is: dQ dt = A (T^4 - T₀^4) The rate of decrease of temperature is given by: dT dt = 1 mc dQ dt = A (T^4 - T₀^4) mc Given values are: = 5.67 10⁻⁸ W m ⁻² K ⁻⁴ T = 227^ C = 500 K T₀ = 27^ C = 300 K m = 1.0 kg c = 400 J kg ⁻¹ K ⁻¹ Substituting these values into the equation: dT dt = 5.67 10⁻⁸ 0.015 (500^4 - 300^4) 1.0 400 dT dt = 5.67 10⁻⁸ 0.