Question

Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper and one of iron. You place the two side by side and braze them together, making a composite bar with a cross-section of 2 cm2. If one end of this rod is placed in boiling water and the other end in ice water, how much power will be conducted through the rod when it reaches steady state?

Answer 1

Explanation:

Expression to calculate thermal resistance for iron ($R_{I}$) is as follows.

             $R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}$  

where,   $L_{I}$ = length of the iron bar

             $k_{I}$ = thermal conductivity of iron

             $A_{I}$ = Area of cross-section for the iron bar

Thermal resistance for copper ($R_{c}$) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where,  $L_{c}$ = length of copper bar

             $k_{c}$ = thermal conductivity of copper

            $A_{c}$ = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

           Q = $\frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

 Putting the given values into the above formula as follows.

       Q = $\frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

  = $\frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}$

           = 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

                 P = $\frac{Q}{T}$

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.