Steam at `100^(@)C` is passed through a tubeof radius 5 cm and length 3 m. If the thickness of tube be 2 mm and conductivity of its material be `2xx10^(-4)" cal "cm^(-1) K^(-1) s^(-1)`, calculate the rate of loss of heat in `js^(-1)` . The outside temperature is `20^(@)C`.

Answer

`[3168.1 js^(-1)]`

Explanation

Here,`Delta T=T_(1)-T_(2)=100-20=80^(@)C, r=5 cm`,
`Delta x=2 mm=0.2" cm, "l=3 m=300 cm`,
`K=2xx10^(-4)" cal "cm^(-1) K^(-1) s^(-1)`
Area `A=2 pi r l=2 pixx5xx300=3 pixx10^(3) cm^(2)`
Rate of loss of heat is
`(dQ)/(dt)=(KA Delta T)/(Delta x)=(2xx10^(-4))xx3pixx10^(3)xx(80)/(0.2)`
`=754.3 cal s^(-1)`
`=754.3xx4.2 js^(-1)=3168.1 js^(-1)`