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1.56xx10^(5) heat energy is transferred through wall 2 m^(2) and 12 cm thickness in every hour. Temperature difference between two walls is 20^(@)C, the mall conductivity of material of wall is. . . .. ."Wm"^(-1)K^(-1) |
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Answer» SOLUTION :Here, `A="2 m"^(2)` `L=12cm=0.12m` `H=1.56xx10^(5)J` `t=1" hour "=3600s` `T_(1)-T_(2)=20^(@)C` or 20 K Heat conducted in time .t. in thermal steady state, `H=(kA(T_(1)-T_(2)))/(L)xxt` `:.k=(HL)/(A(T_(1)-T_(2))xxt)` `=(1.56xx10^(5)xx0.12)/(2xx20xx3600)` `=0.0000013xx10^(5)` `:.k=0.13" W m"^(-1)K^(-1)` |
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