An iron bar `(L_(1)=0.1" m",A_(1)=0.02" m"^(2),K_(1)=97" Wm"^(-1)K^(-1))` and a brass bar `(L_(2)=0.1" m",A_(2)=0.02" m"^(2),K_(2)=109" W m"^(-1)K^(-1))` are soldered end to end as shown in figure. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar, and (iii) the heat current through the compound bar.

Explanation

Length of iron rod `L_(1)=0.1m` Area `A_(1)=0.02m^(2)`
Thermal conductivity `K_(1)=79" Wm"^(-1)K^(-1)`
Heat current `=H_(1)`
Length of brass rod `L_(2)=0.1` m
Area `A_(2)=0.02m^(2)`
Thermal conductivity `K_(2)=109" Wm"^(-1)K^(-1)`
Temp. of heat and of combined rod `T_(1)=373` K
Temp. of cold and of combined rod `T_(2)=273K`
Temp. of junction `T_(0)=?`
(i) In thermal steady state `H_(1)=H_(2)=H`
`:.(K_(1)A_(1)(T_(1)-T_(0)))/(L_(1))=(K_(2)A_(2)(T_(0)-T_(2)))/(L_(2))`
But `A_(1)=A_(2)=A` and `L_(1)=L_(2)=L`
`:.K_(1)(T_(1)-T_(0))=K_(2)(T_(0)-T_(2))`
`:.K_(1)T_(1)-K_(1)T_(0)=K_(2)T_(0)-K_(2)T_(2)`
`:.K_(1)T_(1)+K_(2)T_(2)=T_(0)(K_(1)+K_(2))`
`:.T_(0)=(K_(1)T_(1)+K_(2)T_(2))/(K_(1)+K_(2))`
`=(79xx373+109xx273)/(79+109)`
`=(29467+29757)/(188)`
`=(59224)/(188)`
`:.T_(0)=315K`
(ii) thermal resistance `R_(H)=(2L)/(KA),R_(H_(1))=(L)/(K_(1)A)`
and `R_(H_(2))=(L)/(K_(2)A)`
`:.` In series connection `R_(H)=R_(H_(1))+R_(H_(2))`
`:.(2L)/(KA)=(L)/(K_(1)A)+(L)/(K_(2)A)`
`:.(2)/(K)=(1)/(K_(1))+(1)/(K_(2))`
thermal conductivity of combined rod
`K=(2K_(1)K_(2))/(K_(1)+K_(2))`
`=(2xx79xx109)/(79+109)`
`=(17222)/(188)`
`:.K=91.6" Wm"^(-1)K^(-1)`
`rArr` (iii) Heat current in combined rod,
`H=(KA(T_(1)-T_(2)))/(2L)`
`=91.6xx0.02xx(373-273)/(2xx0.1)`
`:.H=916W`