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Difficulty: HardHeat Transfer: Radiation

In a vacuum testing chamber, two identical engine heat sinks—one with a highly polished aluminum finish and the other painted matte black—are heated to exactly 200°C. True or false: The matte black heat sink will emit thermal radiation at a significantly higher rate than the polished aluminum one.

Answer: Answer

Answer

True
The statement is true because a surface's ability to emit thermal radiation (its emissivity) heavily depends on its color and texture. Dark, matte surfaces are almost ideal emitters of thermal radiation, whereas shiny, polished metals are highly reflective and trap heat internally, making them very poor emitters. Consequently, the matte black heat sink radiates its thermal energy into the vacuum much faster than the polished one.

Step-by-Step Solution

1
Identify the available mode of heat transfer in the described environment.
Because the heat sinks are in a vacuum, thermal energy can only be transferred via radiation. Conduction and convection require a physical medium (like air or liquid) and are not possible here.
Establishing the correct heat transfer mechanism is critical to prevent applying the wrong physical rules.
2
Evaluate the effect of surface finish on thermal radiation emission.
A matte black finish has high emissivity, meaning it is an excellent radiator of thermal energy. A highly polished aluminum finish has low emissivity, making it a very poor radiator.
Emissivity is the key material property that dictates how efficiently a surface emits thermal radiation.
3
Compare the heat emission rates of the two heat sinks.
Since both heat sinks are at the exact same temperature, the matte black heat sink will emit radiant heat at a significantly higher rate than the polished aluminum one.
At any given temperature, an object with a higher emissivity always radiates more thermal power than an object with lower emissivity.

Key Concept

Surface Emissivity and Thermal Radiation
Estimated Time:45s
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Topics

Heat Transfer: Radiation

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