The walls of a closed cubical box of edge 60 ~cm are made of material of thickness 1 ~mm and thermal conductivity, 4 10⁻⁴ cal s ⁻¹ ~cm ⁻¹ ^ C ⁻¹ . The interior of the box is maintained 1000^ C above the outside temperature by a heater placed inside the box and connected across 400 ~V DC supply. The resistance of the heater is

Step-by-step solution

Key Idea Heat transfer through conduction wall is given by mathematical expression, d Q d t =k A (T₁-T₀ ) x where, d Q d t = power transferred through the wall and x= thickness of the wall Here, side of cube, a=60 ~cm . Hence, area A=6 a , Total area =6 (area of one side of the cube) thickness x=0.1 ~cm , thermal conductivity, k=4 10⁻⁴ cal s ⁻¹ ~cm ⁻¹ ^ C ⁻¹ , temperature difference, (T₁-T₀ )=1000^ C and potential of DC source V=400 ~V Hence, the power aligned P & = k A 4.184 (T₁-T₀ ) x P & = k 6 a^2 4.184 10^3 x P