Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. Three metal rods made of copper, a 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.
When steady state condition is reached everywhere, which of the following statements is true?

Options

A. 86°C
B. 18.8°C
C. 57°C
D. 73°C

Answer

A

Explanation

Since heat transmitted 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)andK_(3)` are thermal conductivities of copper, aluminium and brass, respectively and `T_(ca)andT_(ab)` are the steady state temperatures of copper–aluminium and aluminium–brass junctions,
Now , `K_(1)=2K_(2)=4K_(3)` given
Hence, `4(100-T_(ca))=2(T_(ca)-T_(ab))=T_(ab)`
Solving for `T_(ab)andT_(ca)` we obtain , `14T_(ca)=1200`
`rArrT_(ca)=(1200)/(14)=85.7=86^(@)C "Also"T_(ab)=(2)/(3)T_(ca)=(2xx86)/(3)=57^(@)C`