A brass plate with a thickness of 1.00 cm is placed over a copper plate with a thickness of 2.00 cm. Take the thermal conductivity of brass to be 100 W/m °C and that of copper to be 400 W/m °C. What is the effective thermal conductivity for the brass-copper layer?

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Quizplus · Accepted Answer

The effective thermal conductivity ($k_{eff}$) of a series of layers can be calculated using the formula for thermal resistances in series, which is analogous to electrical resistances. The total thermal resistance ($R_{total}$) is the sum of the individual resistances ($R = \frac{L}{kA}$, where $L$ is the thickness, $k$ is the thermal conductivity, and $A$ is the cross-sectional area). Since the area $A$ cancels out and we're looking for an effective conductivity over a fixed distance (the sum of the thicknesses of both materials), we can rearrange the formula to solve for $k_{eff}$ in terms of the thickness ($L$) and conductivity ($k$) of each layer.Given:- Brass: $L_1 = 1.00$ cm, $k_1 = 100$ W/m°C- Copper: $L_2 = 2.00$ cm, $k_2 = 400$ W/m°CThe total thickness $L_{total} = L_1 + L_2 = 3.00$ cm.The total thermal resistance is the sum of the individual resistances:$$R_{total} = \frac{L_1}{k_1} + \frac{L_2}{k_2}$$Substituting the given values:$$R_{total} = \frac{1.00}{100} + \frac{2.00}{400} = 0.01 + 0.005 = 0.015 \, \text{m}^2°C/W$$To find the effective thermal conductivity ($k_{eff}$), we use the total thickness and the inverse of the total resistance, since $k_{eff} = \frac{L_{total}}{R_{total}A}$, and again, $A$ cancels out for our purposes:$$k_{eff} = \frac{L_{total}}{R_{total}} = \frac{3.00}{0.015} = 200 \, \text{W/m}°C$$Therefore, the effective thermal conductivity for the brass-copper layer is 200 W/m°C.

A Brass Plate with a Thickness of 1