Question

Heat flux meters use a very sensitive device known as a thermopile to measure the temperature difference across a thin, heat conducting film made of kapton \((k=0.345 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the thermopile can detect temperature differences of \(0.1^{\circ} \mathrm{C}\) or more and the film thickness is \(2 \mathrm{~mm}\), what is the minimum heat flux this meter can detect? A?swer: \(17.3 \mathrm{~W} / \mathrm{m}^{2}\)

Short answer

Answer: The minimum detectable heat flux for this heat flux meter is 17.3 W/m².

Step-by-step solution

Convert the thickness to meters

Since the given thickness is in millimeters, we need to convert it to meters to maintain consistent units. We can do this by multiplying by the conversion factor (1 meter = 1000 millimeters): Thickness (m) = 2 mm * (1 m / 1000 mm) = 0.002 m

Re-write Fourier's Law with the known variables

Now, we can rewrite Fourier's Law using the given values for k and dT, and the converted value for dx: q = -0.345 W/m·K * (0.1°C / 0.002 m)

Calculate the minimum heat flux

Finally, we can calculate the minimum detectable heat flux by solving for q: q = -0.345 W/m·K * (0.1 K / 0.002 m) = 17.25 W/m² Since negative values indicate the direction of heat flux, we can present the result as a positive value: Minimum detectable heat flux = 17.3 W/m²