Two roads of same length and cross section area A1 and A2 and conductivity K1 and K2 and temperature T1 and T2 are joined find conductivity
suspect you mean rods
you did not say if parallel or end to end
if end to end, add resistances
if parallel add conductances like 1/R = 1/R1 + 1/R2
To determine the effective conductivity of the joined roads, we need to consider the thermal conductivity of each road and the temperature difference between the two roads.
The formula to calculate the effective conductivity is:
K_effective = (K1 * A1 + K2 * A2) / (A1 + A2)
Here, K1 and K2 are the conductivities of the two roads, and A1 and A2 are their respective cross-sectional areas.
Additionally, we need to account for the temperature difference between the two roads, as it affects the overall heat transfer. If T1 and T2 are the temperatures of the two roads, we can calculate the temperature difference as:
ΔT = T1 - T2
By substituting the calculated K_effective value and the temperature difference ΔT into equations relating heat and conductivity, we can ascertain the heat transfer between the two roads.
It's important to note that these equations assume the roads are in thermal equilibrium, meaning there is no heat loss to the surroundings. If there is a significant amount of heat loss, the equations would need to be adjusted accordingly.