Final answer:
The thermal conductivity (k) of the test panel is calculated to be approximately 0.527 W/m·K using the steady-state heat conduction equation derived from Fourier's law.
Step-by-step explanation:
To find the thermal conductivity (k) of the test panel, we need to use the steady-state heat conduction equation, which is derived from Fourier's law of heat conduction:
Q = k × A × (ΔT / d)
Where:
- Q is the rate of heat transfer in watts (W)
- k is the thermal conductivity in Watts per meter kelvin (W/m·K)
- A is the area through which heat is being transferred in square meters (m²)
- ΔT is the temperature difference across the material in kelvin (K) or degrees Celsius (°C)
- d is the thickness of the material in meters (m)
First, we must convert all given units to their SI equivalents:
- The thickness of the test panel, d = 2.54 cm = 0.0254 m
- The area, A = 20.32 cm × 20.32 cm = 0.2032 m × 0.2032 m = 0.04129 m²
- The temperature difference, ΔT = 79.4°C - 21.1°C = 58.3°C
- The power, Q = 50 W (already in SI units)
Plugging these values into the equation, we get:
50 W = k × 0.04129 m² × (58.3 / 0.0254 m)
Then, solving for k:
k = 50 W / (0.04129 m² × (58.3 / 0.0254 m))
Cancelling the units and calculating:
k = 50 / (0.04129 × 2296.85)(W/(m²·K))
k = 50 / 94.827 (W/m·K)
k ≈ 0.527 W/m·K
The thermal conductivity (k) of the test panel is approximately 0.527 W/m·K.