AIIMS 2006 Physics Heat and Temperature MCQ Question

To solve the question of which object glows brightest when heated to 2000°C, we need to understand the concept of thermal radiation and how it relates to the color of the objects.

1. Explanation of the Correct Answer

The brightness of an object when heated is determined by its emissivity, which is a measure of how effectively a surface emits thermal radiation. Emissivity values range from 0 to 1, where 1 indicates a perfect black body that emits the maximum amount of thermal radiation for a given temperature.

In this case, we have three colored objects: black, gray, and white. The properties of these colors in terms of thermal radiation are as follows:

• Black Object: It has the highest emissivity (close to 1), meaning it is excellent at emitting thermal radiation.
• Gray Object: It has an emissivity value less than that of the black object, but greater than that of the white object.
• White Object: It has the lowest emissivity (close to 0.5), meaning it is less effective at emitting thermal radiation.

However, the question states that all three objects are thrown into a furnace and reach the same temperature of 2000°C. According to Planck's Law for black body radiation, the intensity of the emitted radiation (brightness) is determined by the temperature of the object.

The formula for the power emitted per unit area of a black body is given by the Stefan-Boltzmann Law:

$$E = \sigma T^4$$

where:

• $E$ is the emissive power (brightness),
• $\sigma$ is the Stefan-Boltzmann constant ($5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4$),
• $T$ is the absolute temperature in Kelvin.

Since all three objects are at the same temperature of 2000°C, we can convert this to Kelvin:

$$T = 2000°C + 273.15 = 2273.15 \, K$$

Now, substituting $T$ into the Stefan-Boltzmann Law, we see that the emitted power per unit area will be the same for all three objects:

$$E = \sigma (2273.15)^4$$

Since the temperature is the same, the emitted brightness in terms of power per unit area will also be the same, regardless of the color of the objects. Therefore, all three objects will glow with equal brightness at 2000°C.

2. Clarification of the Incorrect Options

• Option A (the white object): Although the white object has the lowest emissivity and emits less thermal radiation compared to the black and gray objects, it still reaches the same temperature as the others. Thus, it does not glow the brightest, and all objects glow equally.

• Option B (the black object): While the black object has the highest emissivity, at the same temperature as the others, it does not glow brighter than the others in this scenario. All emit the same amount of power per unit area.

• Option D (the gray object): Similar to the black object, the gray object has a higher emissivity than the white one but lower than the black. However, all three are at the same temperature, resulting in equal brightness.

Conclusion

In summary, when heated to the same temperature, the brightness of the objects is solely dependent on their temperature, not their color. Therefore, the correct answer is C) all glow with equal brightness. All three objects, regardless of their color, emit the same amount of thermal radiation at 2000°C due to their equal temperature.