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In example 3, K(1) = 0.125 W//m-.^(@)C, ...

In example `3, K_(1) = 0.125 W//m-.^(@)C, K_(2) = 5K_(1) = 0.625 W//m-.^(@)C` and thermal conductivity of the unknown material is `K = 0.25 W//m^(@)C L_(1) = 4cm, L_(2) = 5L_(1) = 20cm`. If the house consists of a single room of total wall area of `100 m^(2)`. then find the power of the electric heater being used in the room.

Text Solution

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`ul(I^(ulst)"method") R_(1) =R_(2) = ((4xx10^(-2)m))/((0.125w//m-^(@)C)(100m^(2))) =32 xx 10^(-4).^(@)C//w`
`:. (25 -20)/(R_(1)) = (20-T_(3))/(R )`
`R = (L)/(KA) = 112 xx 10^(-4).^(@)C//W`
the equivalent thermal resistance of the entire wall `= R_(1) +R_(2) +2R = 288 xx 10^(-4).^(@)C//W`
`:.` Net heat current i.e. amount of heat flowing out of the house per second `= (T_(H)-T_(C))/(R)`
`=(25^(@)C-(-20^(@)C))/(288xx10^(4).^(@)C//w) =(45xx10^(4))/(288)`watt
`= 1.56 K`watt
Hence the heater must supply `1.56 kW` to compenstate for the outflow of heat.
`ul(II^(ulnd)method)`
`i = (T_(1)-T_(2))/(R_(1)) = (25 -20)/(32 xx 10^(-4)) = 1.56 K`watt
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