Explanation:
Expression to calculate thermal resistance for iron (
) is as follows.
where,
= length of the iron bar
= thermal conductivity of iron
= Area of cross-section for the iron bar
Thermal resistance for copper (
) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]
where,
= length of copper bar
= thermal conductivity of copper
= Area of cross-section for the copper bar
Now, expression for the transfer of heat per unit cell is as follows.
Q = 
Putting the given values into the above formula as follows.
Q = 
= 
= 2.92 Joule
It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,
P = 
Here, T is 1 second so, power conducted is equal to heat transferred.
So, P = 2.92 watt
Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.