A steel bar 10.0 cm long is welded end to end to a copper bar 20.0 cm long. Both bars are insulated perfectly on their sides. Each bar has a square cross-section, 20.00 cm on a side. The free end of the steel bar is maintained at `100^(@)C` by placing it in contact with steam, and the free end of the copper bar is maintained at `0^(@)C` by placing it in contact with ice. find the temperature at the junction of the two bars and the total rate of heat flow.

Explanation

In the steady state, the two heat currents are equal. We have ltBrgt `H_("steel")=(K_("steel")A(100^(@)C-T))/(L_("steel"))and H_("copper")=(K_("copper")A(T-0^(@)C))/(L_("copper"))`
The areas A are equal an may be cancelled out.
Substituting `L_("steel")=0.100m,L_("copper")=0.200m`, and numerical values of K from the previous table, we find
`((50.2W//m.K)(100^(@)C-T))/(0.100m)=((385W//m.K)(T-0^(@)C))/(0.200m)`
Solving for T, we obtain `T=20.7^(@)C` ltBrgt We can find the total heat current by substituting this value for T,
`H_("steel")=((50.2W//m.K)(0.0200m^(2))(100^(@)C-20.7^(@)C))/(0.100m)`
`becauseH_("steel")=15.9W`
`thereforeH_("copper")=15.9W`