Problem 83 - Master Resource Book in JEE Main Physics

Question

A black body at a temperature of \(327^{\circ} \mathrm{C}\) radiates \(4 \mathrm{cal} \mathrm{cm}^{-2} \mathrm{~s}^{-1}\), At a temperature of \(927^{\circ} \mathrm{C}\), the rate of heat radiated per unit area in cal \(\mathrm{cm}^{-2} \mathrm{~g}^{-1}\) will be (a) 16 (b) 32 (c) 64 (d) 128

Step-by-Step Solution

Answer

The rate of heat radiated at 927°C is 64 cal cm⁻² s⁻¹.

Step 1: Convert Temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. Recall that Kelvin is given by the formula \( T(K) = T(°C) + 273 \).For \(327^{\circ} \text{C}\), the temperature in Kelvin is:\[ T_1 = 327 + 273 = 600 \text{ K} \]For \(927^{\circ} \text{C}\), the temperature in Kelvin is:\[ T_2 = 927 + 273 = 1200 \text{ K} \]

Step 2: Apply Stefan-Boltzmann Law

The Stefan-Boltzmann Law relates the power radiated by a black body to its temperature by the equation:\[ E \propto T^4 \]Given that the energy radiated at \(327^{\circ} \text{C}\) is \(4 \text{ cal cm}^{-2} \text{s}^{-1}\), we can find the new energy radiated at \(927^{\circ} \text{C}\).

Step 3: Calculate the Rate of Heat Radiated

Use the proportional relationship from Stefan-Boltzmann Law:\[ \frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 \]Substitute the known values:\[ \frac{E_2}{4} = \left(\frac{1200}{600}\right)^4 = 2^4 = 16 \]Thus, the rate of heat radiated \(E_2\) is:\[ E_2 = 4 \times 16 = 64 \text{ cal cm}^{-2} \text{s}^{-1} \]

Key Concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in physics that helps us understand how objects emit heat. This law states that the total energy radiated per unit area of a black body is directly proportional to the fourth power of its temperature in Kelvin. In simpler terms, as the temperature of an object increases, the amount of heat it radiates grows rapidly.To express this mathematically, we use the formula: \[ E ext{ (energy radiated)} \ ext{ is proportional to } \ T^4 \]where \( E \) represents the energy per unit area, and \( T \) is the temperature in Kelvin.Black bodies are idealized objects that absorb all incoming radiation and emit energy perfectly. Stefan-Boltzmann Law is critical for calculating the radiation emitted by stars, planets, and other astronomical bodies. It highlights how significant changes in temperature lead to dramatic increases in the heat emitted. For example, doubling the temperature would result in a heat emission that is 16 times greater. This concept is essential for understanding how temperature variations affect the behavior of radiating objects.

Temperature Conversion

Converting temperatures is often one of the first steps when dealing with problems involving black body radiation. In our exercise, we need to convert temperatures from Celsius to Kelvin since the Kelvin scale is used in scientific formulas, including the Stefan-Boltzmann Law.The conversion formula is simple:\[ T(K) = T(°C) + 273 \]where \( T(K) \) is the temperature in Kelvin and \( T(°C) \) is the temperature in Celsius. This formula reflects that the Kelvin scale starts at absolute zero, which is –273°C. Thus, to convert 327°C to Kelvin, we add 273, resulting in 600 K. Similarly, 927°C becomes 1200 K upon conversion.Kelvin is the SI unit for temperature and is used because it facilitates direct comparisons of thermal energy. So, utilizing the Kelvin scale is a common practice in scientific calculations to align with laws involving thermodynamics and radiation, ensuring accuracy.

Radiated Heat Calculation

Calculating the rate of heat radiated is the next step after understanding the Stefan-Boltzmann Law and converting temperatures appropriately. The law shows us the relationship between temperature and energy emitted by a black body, allowing us to predict how much energy an object will radiate at a given temperature.For example, using the ratios established by the Stefan-Boltzmann Law:\[ \frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 \]where \( E_1 \) is the initial energy emission, \( E_2 \) is the new energy emission that we want to find, and \( T_1 \) and \( T_2 \) are the initial and new temperatures respectively.Substituting known values provides:\[ \frac{E_2}{4} = \left(\frac{1200}{600}\right)^4 = 2^4 = 16 \]Therefore, \( E_2 = 4 \times 16 = 64 \text{ cal cm}^{-2} \text{s}^{-1} \).Such calculations help quantify how much more energy is being radiated as the temperature increases, making it a valuable tool in various scientific and engineering applications. Understanding how to perform these calculations enables one to better understand energy changes in thermal environments.