Steam at 373 `K` is passed through a tube of radius `10 cm` and length` 10 cm` and length `2m`. The thickness of the tube is `5mm` and thermal conductivity of the material is` 390 W m^(-1) K^(-1)`, calculate the heat lost per second. The outside temperature is `0^(@)C`.

Explanation

Using the relation `Q = (KA (T_1-T_2)t)/L`
Here, heat is lost through the cylindrical surface of the tube.
`A = 2pi` (radius of the tube) (length of the tube)
`= 2pi xx 10 cm xx 2m`
`= 2pi xx 0.1 xx2m^2 = 0.4 pi m^2`
`K = 390 W m^(-1) K^(-1)`
`T_(1) = 373 K , T_(2) = 0^(@) C = 273 K`
`L = 5 mm = 0.005 m and t = 1 s `
`therefore Q = (390 W m^(-1) xx 0.4 pi m^(2) xx (373 - 273) K xx 1 s)/(0.005 m)`
`= (390 xx 0.4 pi xx 100)/(0.005) Ws = 98 xx 10^(5) J`