An experiment is performed to determine the emissivity of a certain surface coat

Question

An experiment is performed to determine the emissivity of a certain surface coating. For this purpose, two identical small cubes of copper are prepared. One of the cubes is painted white ( epsilon = 0. 954). The other is also coated, but the epsilon of that coating is unknown. Heaters are embedded in each cube to control the respective surface temperatures. The copper provides an identical uniform temperature on all surfaces. The cubes are placed in a very large enclosure whose walls are at 70 F, and the air in the enclosure is also at 70 F. The two cubes are positioned sufficiently far apart so that they do not interact with each other, either radiatively or convectively. When the experiment is operated so that both cubes have the same surface temperature of 122 F, the respective heat transfer rates for the cubes are: (a) for the white-painted cube, Q / A = 112. 3 Btu / hr-ft2 and (b) for the other cube, Q / A = 94. 3 Btu / hr-ft2. Required result: Find the epsilon value of the other cube.

Explanation / Answer

WE know heat transfer rate due to radiation =(T^4 - t^4)

where T and t are in Kelvin ,

here all other parameters are same , so the difference in heat transfer is due to emmisivity only.

so H1/H2= 1/2

112.3 / 94.3 = 0.954/2

so 2 = 0.801

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