Problem 75 - Master Resource Book in JEE Main Physics

Question

Two bodies \(A\) and \(B\) are placed in an evacuated vessel maintained at a temperature of \(27^{\circ} \mathrm{C}\). The temperature of \(A\) is \(327^{\circ} \mathrm{C}\) and that of \(B\) is \(227^{\circ} \mathrm{C}\). The ratio of heat loss from \(A\) and \(B\) is about (a) \(2: 1\) (b) \(4: 1\) (c) \(1: 2\) (d) \(1: 4\)

Step-by-Step Solution

Answer

The ratio of heat loss from A and B is approximately 2:1.

Step 1: Understand the Concept of Heat Loss

Heat loss from a body can be calculated using Stefan-Boltzmann Law in the form of \( E \propto (T^4 - T_0^4) \), where \( T \) is the temperature of the body and \( T_0 \) is the surrounding temperature. To compare the heat loss from two bodies, this expression can be used to compare their respective heat losses.

Step 2: Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula \( T(K) = T(^{\circ}C) + 273 \). We have:- Temperature of \(A\): \(327^{\circ}C + 273 = 600\,K\)- Temperature of \(B\): \(227^{\circ}C + 273 = 500\,K\)- Surrounding temperature: \(27^{\circ}C + 273 = 300\,K\)

Step 3: Calculate the Effective Temperature for Heat Loss

Use the formula from Step 1, \( E_A \propto (600^4 - 300^4) \) and \( E_B \propto (500^4 - 300^4) \), where \( E_A \) and \( E_B \) are the emissive powers or rate of heat loss from bodies \( A \) and \( B \) respectively.

Step 4: Simplify the Power of Four

Calculate the exponent values:\( 600^4 = (6 \times 10^2)^4 = 129600000000 \) and \( 300^4 = (3 \times 10^2)^4 = 8100000000 \)So, \( E_A \propto (129600000000 - 8100000000) = 121500000000 \)Similarly,\( 500^4 = (5 \times 10^2)^4 = 62500000000 \)So, \( E_B \propto (62500000000 - 8100000000) = 54450000000 \)

Step 5: Calculate the Ratio of Heat Loss

Now find the ratio \( \frac{E_A}{E_B} = \frac{121500000000}{54450000000} \).Divide numerator and denominator by \( 54450000000 \):- \( \frac{121500000000}{54450000000} \approx \frac{2}{1} \).

Step 6: Choose the Correct Option

From the calculation, the ratio of heat loss \( \frac{E_A}{E_B} \) is \( 2:1 \). Hence, the correct option is \( (a) \ 2:1 \).

Key Concepts

Understanding Heat Loss with Stefan-Boltzmann Law

Heat loss is a term used to describe how heat energy leaves a body. When a body is hotter than its surroundings, it loses heat to the environment. This concept is quantified using the Stefan-Boltzmann Law. According to this law, the emissive power of a body, or its power to radiate heat, is proportional to the fourth power of its temperature in Kelvin, minus the fourth power of the surrounding temperature, also in Kelvin. This relationship is mathematically expressed as:\[ E \propto (T^4 - T_0^4) \]where:

• \( E \) represents the emissive power or rate of heat loss.
• \( T \) is the temperature of the body.
• \( T_0 \) is the temperature of the surroundings.

Understanding this formula is crucial for analyzing how bodies at different temperatures lose heat when placed in a common environment. This law applies mostly to ideal conditions where bodies and surroundings are perfect emitters or absorbers of radiation.

Temperature Conversion: Celsius to Kelvin

In scientific calculations, particularly those involving thermal physics, it is crucial to convert temperatures from Celsius to Kelvin. The Kelvin scale is an absolute temperature scale, which is why it's preferred in scientific equations involving heat and temperature.The conversion formula is simple:\[ T(K) = T(^{\circ}C) + 273.15 \]For rounding convenience, we often use 273. This ensures that all temperatures are in the same unit, which makes it easier to apply mathematical formulas.For example, in our exercise, the given temperatures are converted as follows:- Temperature of body \(A\): from \(327^{\circ}C\) to \(600\, K\).- Temperature of body \(B\): from \(227^{\circ}C\) to \(500\, K\).- Surrounding environment: from \(27^{\circ}C\) to \(300\, K\).Converting to Kelvin ensures we are accurately using the Stefan-Boltzmann Law for heat loss computations.

Emissive Power and its Role in Heat Loss

Emissive power is a term describing how effectively a body radiates heat energy. In the context of the Stefan-Boltzmann Law, it represents the rate at which heat is lost from a body to its surroundings. The higher the emissive power, the greater the rate of heat loss from the body.The calculation of emissive power is derived from the temperature difference between the body and its surroundings, specifically using their temperatures raised to the fourth power. This highlights the exponential nature of heat loss relative to temperature:- For body \(A\), the calculation is: \[ E_A \propto (600^4 - 300^4) \]- For body \(B\), it is: \[ E_B \propto (500^4 - 300^4) \]After the computations, we find the heat loss from each body can be compared by taking the ratio of their emissive powers. For two bodies at differing temperatures in the same environment, the body with a higher temperature will typically have a higher emissive power, meaning it loses heat more rapidly.This explains why, in our exercise, the rate of heat loss from body \(A\) was calculated to be twice as much as from body \(B\), leading to the ratio \(2:1\). Understanding emissive power helps us grasp why different bodies, even in the same environment, can lose heat at very different rates.