Question

(23 - 27) a 3/16 in. thick acrylic testing box with dimensions of 10.0 in. x 10.0 in. is covered with an unknown 0.50 in. insulation material. determine the thermal conductivity for the insulating material if a 25.0w bulb is used to heat the box. the bulb maintains the inside temperature at 10.0°c higher than the outside temperature.

1. list all known values.
2. list all unknown values.
3. select equations.
4. apply known values.
5. solve.

Explanation:

23. List all known values

• Thickness of acrylic box: \( \frac{3}{16}\text{ in}\) (but not needed for calculating thermal conductivity of insulation).
• Dimensions of the box (for area calculation): \(10.0\text{ in}\times10.0\text{ in}\).
• Thickness of insulation material \(L = 0.50\text{ in}\).
• Power of the bulb (heat transfer rate) \(Q/t=25.0\text{ W}\).
• Temperature difference \(\Delta T = 10.0^{\circ}\text{C}\).

The problem asks to determine the thermal conductivity of the insulating material. Thermal conductivity is denoted by \(k\).

The formula for heat transfer by conduction is \( \frac{Q}{t}=\frac{kA\Delta T}{L}\), where \(\frac{Q}{t}\) is the heat transfer rate, \(k\) is the thermal conductivity, \(A\) is the area, \(\Delta T\) is the temperature difference, and \(L\) is the thickness of the material.

Answer:

Known values:

• Area \(A=(10.0\text{ in})\times(10.0\text{ in}) = 100\text{ in}^2\). Convert to \(m^2\): \(1\text{ in}^2 = 6.4516\times10^{-4}\text{ m}^2\), so \(A = 100\times6.4516\times10^{-4}\text{ m}^2=0.064516\text{ m}^2\).
• Thickness of insulation \(L = 0.50\text{ in}\). Convert to \(m\): \(1\text{ in}=0.0254\text{ m}\), so \(L = 0.50\times0.0254\text{ m}=0.0127\text{ m}\).
• Heat transfer rate \(\frac{Q}{t}=25.0\text{ W}\).
• Temperature difference \(\Delta T = 10.0^{\circ}\text{C}\).

24. List all unknown values