Question

A copper rod and an aluminum rod of equal diameter are joined end to end in good thermal contact. The temperature of the free end of the copper rod is held constant at 100°C, and that of the far end of the aluminum rod is held at 0°C. If the copper rod is LCu = 0.698 m long, what must be the length of the aluminum rod so that the temperature at the junction is 50.0°C?

Answer 1

Answer:

The length of aluminium is 0.41253 m.

Explanation:

According to the question.

i) Diameter of copper rod = diameter of aluminium rod

ii) The temperature at the free end of copper  = 100°C.

iii) The temperature of aluminium at free end = 0°C.

iv) Length of copper rod = 0.698 m.

v) The temperature at the junction = 50°C.

Now we have to find the length of aluminium rod.

Let the length of aluminium rod is 'x'.

By using the Fourier law of heat conduction.

$Q = K*A *\frac{(T_{h}-T_{l} ) }{L}$

K = thermal conductivity.

A = area.

Q = energy transfer rate.

L = length.

$T_{h}$ = higher temperature.

$T_{l}$ = lower temperature.

Since the temperature are held constant energy transfer rate ( Q ) through copper rod is equal to the energy transfer rate ( Q ) through aluminium rod.

$Q_{copper} = Q_{aluminium}$

$K_{copper} *A *\frac{(100-50 ) }{L_{copper} } = K_{aluminium} *A *\frac{(50-0) }{L_{aluminium} }$

Area and temperature difference gets cancelled.

$L_{aluminium} = \frac{L_{copper}*K_{aluminium} }{K_{copper} }$

$K_{copper$ = 401 k ( w/(m°C))

$K_{aluminium$ = 237  ( w/(m°C))

$L_{aluminium} = \frac{0.698*237 }{401 }$    = 0.41253 m.