If a very hot object has emissivity less than 1.0, will its apparent temperature be lower, higher, or the same as the true temperature?

• A. Higher
• B. The Same
• C. Lower
• D. Cannot Be Determined

Answer: (C)

Emissivity is how effectively an object emits radiation compared with a perfect blackbody. If an object is hotter than its surroundings but has emissivity less than 1, it emits less radiant power than a blackbody at the same temperature. When a detector translates the measured radiance into a temperature assuming emissivity is 1, it infers a temperature lower than the object's true temperature. In formula form, the emitted radiance is proportional to ε T^4. If you interpret that radiance as coming from a perfect emitter, you solve for T_app with ε = 1, giving T_app = ε^(1/4) T. Since ε^(1/4) < 1 for any ε < 1, T_app is smaller than T. For instance, with ε = 0.5, the apparent temperature would be noticeably lower than the true temperature. So the apparent temperature is lower.