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The walls of a closed cubical box of edge `50cm` are made of a material of thickness `1mm` and thermal conductivity `4xx10^(-4) cm^(-1)(.^@C)^(-1)`. The interior of the box is maintained `100^(@)C` above the outside temperature by a heater placed inside the box and connected across `420V` d.c. calculate the resistance of copper. |
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Answer» `A` = total surface area of box `= 6xx` area of each face `= 6xx50xx50xx10^(-4)m^(2) = 1.5m^(2)` `K = 4xx4.2xx10^(-2)J//sm.^(@)C` `theta_(2)-theta_(1) = 100^(@)C, L = 1 mm = 10^(-3)m` The rate of heat transfer through walls `P = Q/t = (KA(theta_(2)-theta_(1)))/(L) = (4xx4.2xx10^(-2)xx1.5xx100)/(10^(-3))` `=25200W` Heat produced by `"heater"//"sec" = (V^2)/(R)` `P = (V^2)/(R)` `R = (V^2)/(P) = (420xx420)/((4xx4.2xx10^(-2)xx1.5xx100)//10^(-3))` `=7Omega`. |
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