Heat transfer equation puzzling results

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SUMMARY

The discussion focuses on calculating the heat transfer rate using the heat transfer equation for a heated plastic plate in air. The formula applied is Q = k × A × ΔT / d, where k is the thermal conductivity of plastic (0.2 W/(mK)), A is the area (0.15 m x 0.15 m), ΔT is the temperature difference (1 K), and d is the thickness of the plastic (0.002 m). The calculated heat transfer rate is 2.25 W, which raises concerns about its reasonableness, especially when considering the low conductivity of air, which is approximately 0.024 W/(mK). The discussion highlights the importance of including convection and radiation effects in heat transfer calculations.

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Using heat transfer equation to find out heat transfer rate but reached a puzzling result, where did I do wrong?

Problem statement:
Heat transfer rate of a heated plastic plate in air, considering only one side, the size of the plate is 15cm X 15cm, assuming temperature difference is 1K.

Formula to use:
Q = k × A × ΔT / d

Q: heat transfer rate in Watts
k: heat conductivity of plastic, 0.2 W/(mK)
A: area, 15cm x 15cm
ΔT: temperature difference, 1K
d: thickness of plastic, 2mm

Results:
Pluging in the numbers, Q=2.25W. For 1K temperature diff. the palm size plastic can transfer 2W?

The result is not reasonable. Since for 10K temperature diff. that palm size plastic plate can transfer 22.5W of power. How to include the low conductivity of air into calculation?

Thank you very much for your help.
 
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Your value for k, about 0.2 watts per meter per deg. K, is a reasonable value.
Use area = 0.15 meters x 0.15 meters
thickness d = 0.002 meters
ΔT = 1 K
Q = 2.25 watts

Air is very different, and depends on whether there is convection or radiation. This table gives k = 0.024 for air: http://hyperphysics.phy-astr.gsu.edu/hbase/tables/thrcn.html#c1
 

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