As shown in the figure, an aluminium rod and a copper rod, each having a length of 1.0 m and a cross-sectional area of 1 cm^2 , are welded together. One end of the combination is maintained at 20^ ^ ^ C while the other

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Quantrex Academy · Accepted Answer

Correct answer: B. 2.36 J

From the given figure, the aluminium and copper rods are connected in parallel.

The total heat current (heat taken out per second) is the sum of the heat currents through each rod:

H = H_{\text{Al}} + H_{\text{Cu}}

Using the formula for heat conduction,

H = \dfrac{KA\Delta T}{L}

:

H = \dfrac{K_{\text{Al}} A \Delta T}{L} + \dfrac{K_{\text{Cu}} A \Delta T}{L}

H = \dfrac{(K_{\text{Al}} + K_{\text{Cu}}) A \Delta T}{L}

Given values are:

K_{\text{Al}} = 200 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}

K_{\text{Cu}} = 390 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}

A = 1 \text{ cm}^2 = 10^{-4} \text{ m}^2

\Delta T = 60^{}^{{}^{\circ}}\text{C} - 20^{}^{{}^{\circ}}\text{C} = 40^{}^{{}^{\circ}}\text{C}

L = 1.0 \text{ m}

Substituting the values into the equation:

H = \dfrac{(200 + 390) \times 10^{-4} \times 40}{1.0}

H = 590 \times 40 \times 10^{-4}

H = 23600 \times 10^{-4}

H = 2.36 \text{ J s}^{-1}

Therefore, the amount of heat taken out per second is

2.36 \text{ J}

.

As shown in the figure, an aluminium rod and a copper rod, each...

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