Temperatures Interchanged

From the first experiment we can conclude that rod A is more conducting than rod B because the temperature difference between rod $A$ and junction $(= 30^\circ C)$ is less than that of rod $B$ and junction $(= 70^\circ C)$ . Or in other words rod B offers more resistance to the heat flow than rod A . So, in the second experiment when the heat flows from $B \rightarrow A$ the temperature of the junction will be less than $70^\circ C$ due to more resistance from rod $B$ .

I found a solution by making an analogy with simple electric circuits. Think of temperature difference being like an electrical potential difference, and the flow of heat being like an electrical current. I also supposed that there is such a thing as thermal resistance, analogous to electrical resistance. The equivalent of ohms law is then

$\text{flow of heat between two points}=\frac{\text{temperature difference between the points}}{\text{thermal resistance}}$

Now for both the illustrated cases the heat flowing into the junction must equal the heat flowing away from it. So the heat flow in each half of the bimetallic rods must be equal. So for the top rod:

$\frac{30}{R_A}=\frac{70}{R_B}\\ \implies \frac{R_B}{R_A}=\frac{70}{30}$

and for the bottom rod

$\frac{T}{R_A}=\frac{100-T}{R_B}\\ \implies \frac{R_B}{R_A}=\frac{100-T}{T}$

So we can say

$\frac{70}{30}=\frac{100-T}{T}$

This linear equation in T (which can be immediately solved by inspection!) yields

$T=30$

and so $\boxed{\text{T < 100}}$

In all honesty, I am not good at these kinds of questions but I have started to understand. If we think about it in the most simple way possible, we see that the junction is not the two temperature's average, and instead, we see that the temperature of the junction is closer to Rod A, by a difference of 30 degrees, and differs from B by 70 degrees.

My mind instantly went to flipping the temperatures around, making the junction temperature the same difference from ends A and B, and lo and behold, it gives 30 degrees, which is less than 70 degrees.

Someone who is able to see the intricacies in the question, I ask of you to help me understand any flaws in my logic, I can see that this is the case of the 4x4 square, perimeter/area controversy.

I will give a very simple explanation..

A is holding a greater amount of heat or is having a higher heat capacity, or B is having lesser heat capacity or will hold less amount of heat

So in the 2nd case b will lose heat rapidly and the temperature T will become lesser than 70 degrees.

Rate of heat flow in the first case yields the ratio of thermal conductivities of A and B to be equal to 7/3 . In the second case, rate of heat flow yields the junction temperatue equal to 30 degree Celsius, which is less than 70 degree Celsius.

In the first scenario, it is shown that rod A meeting rod B (A = 100°C, B = 0°C) and the temperature at the point of intersection is 70°C. The average of these temperatures would be 50°C if the rods were the same material, but it is indeed 70°C meaning that rod A must have a higher temperature resistance (or heat capacity) than rod B, which has a lower heat resistance/capacity. This results in the temperature being closer to rod A's initial temperature.

Therefore, T must have the same properties meaning that the temperature would be closer to rod A's initial temperature (0°C) and at least lower than 50°C, which is lower than 70°C.

There is fomula thatQ=CmΔT. Q is heat absorbed or released. C is specific heat capacity. ΔT is the change of temepature. Frist,Qa always equil to Qb. In the first experiment,we can know that ΔTa1 is 30 ℃,and ΔTb1 is 70℃. so,70Ca=30Cb. Then,in the second experiment, ΔTa2\ΔTb2=30\70

and ΔTa2+ΔTb2=100℃, so ΔTa2 is 30℃,and the junction is 30℃,of couse less than 70℃