Thermal conductivity question?

. Two rods of equal lengths and cross-sectional areas but of different materials are placed in thermal contact as shown in Figure 2. The thermal conductivity of Q Figure 2


is half that of P. The outer end of P


is at 0o C and that of Q is at 100C.


What is the temperature of the interface at steady state?





A 23.88o C


B 33.33C


C 66.66o C


D 80.39o C

Thermal conductivity question?
There are a lot of assumptions here that you h ave not stated. I'm assuming this is a simple Fourier's Law problem where no heat loss to environment is an assumption.





Fourier's Law is Q=U*A*delta(T)





where U is the conductance of each material, A= surface area, delta T is temperature difference between T(0) and T(steady state).





For this problem steady state means Q(q) = Q(p) and from your problem description we know the U(q)=0.5 U(p) and T(q0)=100 and T(p0)=0 and A(q)=A(p).From this knowledge it becomes a plug and chug equation as below.





U(q)*A(q)*(T(q0)-T(steadystate))= U(p)*A(p)*(T(steadystate)-T(p0))





Plug in what we know and cross out the A(q) and Q(p) since they are equivalent and we come up with...





U(q)*(100-T(steadystate))=U(p)*(T(stea...





substitute U(q)=0.5*U(p) as stated in the problem and we get...





(0.5*U(p))*(100-T(steadystate))=U(p)*(...





Divide through the U(p) and the equation simplifies to





50-(0.5*T(steadystate))=T(steadystate)





move the T(steadystate) to one side...





50 = 1.5(T(steadystate))





T(steadystate) = 33.33 C





Good luck.