A brass plate of thickness 2 cm is in contact with a glass sheet. The exposed surface of the brass is at 80C and that of the glass is at 20C. If the thermal conductivity of brass is 105 W/Km and that

To find the thickness of the glass, we use the principle of steady-state heat conduction, where the rate of heat transfer through the brass plate is equal to the rate of heat transfer through the glass sheet. The formula for the rate of heat transfer ($Q$) is given by:

$$Q = \frac{K A \Delta T}{L}$$

where $K$ is the thermal conductivity, $A$ is the cross-sectional area, $\Delta T$ is the temperature difference, and $L$ is the thickness.

Given:
For brass:

• $L_{brass} = 2 \text{ cm} = 0.02 \text{ m}$
• $K_{brass} = 105 \text{ W/Km}$
• $T_{brass\_exposed} = 80^\circ\text{C}$
• $T_{interface} = 65^\circ\text{C}$

For glass:

• $K_{glass} = 0.8 \text{ W/Km}$
• $T_{glass\_exposed} = 20^\circ\text{C}$
• $T_{interface} = 65^\circ\text{C}$

Both have the same area, $A$.