One end of a thick copper rod is immersed into a steam chamber and the other end h:as metal coiled tube through which cold water at `15^(@)`C is circulated. · If the length of the rod is 0.30 m, thickness 0.04 m, K of copper is 375 `Wm^(-1) K^(-1)` and specific heat of water 4200 `Jkg^(-1)K^(-1)`, then calculate the final temperature of 0.10 kg of water at the end of 10 minutes, the temperatllre at the other end being `80^(@)`C.

Explanation

K=375W`m^(-1)K^(-1),t=10xx60=60s`
`A=(3.142xx(0.04)^2)/4=12.56xx10^(-4)m^2`
`d=0.30m,theta_1=100^(@)C,theta_2=80^(@)C,theta_3=15^(@)C,theta_4`=?
Applying `Q=(KA(theta_1-theta_2)t)/(d)=ms(theta_4-theta_3)`
`theta_4-theta_3=(KA(theta_1theta_2)t)/(msd),theta_1-theta_2=100-80=20^(@)C`
`=(375xx12.568xx10^(-4)xx20xx600)/(0.1xx4200xx0.30)=(565.56)/(126)`
`theta_4-theta_3=4.49^(@)C`
hence `theta_4=theta_3+4.49^(@)=15+4.49^(@)`
`theta_4=19.49^(@)`C