Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. Three metal rods made of copper, aluminium and brass are each 15 cm long and 2 cm in diameter. These rods are placed end to end, with aluminium between the other two. The free ends of the copper and brass rods are maintained at `100^@C` and `0^@C` respectively. The system is allowed to reach the steady state condition. Assume there is no loss of heat anywhere.
Under steady state condition the equilibrium temperature of the copper aluminium junction will be

Options

A. `86^@C`
B. `18.8^@C`
C. `57^@C`
D. `73^@C`

Answer

A

Explanation

Since heat tranmitted per second in the steady state is
`Q=(KA(T_2-T_1))/(L)` and the dimensions of copper, aluminium and brass rods are identical, we must have
`K_1(100-T_(ca))=K_2(T_(ca)-T_(ab))=K_3(T_(ab)-0)`
Where `K_1`,`K_2` and `K_3` are thermal conductivities of copper, aluminium and brass respectively, and `T_(ca)` and `T_(ab)` are the steady state temperature of copper aluminium and aluminium brass junction, respectively.
Now, `K_1=2K_2=4K_3`
Hence `4(100-T_(ca))=2(T_(ca)-T_(ab))=T_(ab)`
Solving for `T_(ab)` and `T_(ca)` we obtain,
`14T_(ca)=1200`
`impliesT_(ca)=(1200)/(14)=85.7=86^@C`
Also `T_(ab)=(2)/(3)T_(ca)=(2xx86)/(3)=57^@C`