The Question is from neet-physics thermodynamics , and it was created by the author ABDUL JILANI BASHA

Question

A metallic sphere with a radius of $r$ and emissivity $e$ is heated to a temperature $T$. If the absolute temperature of the surroundings is $T_0$, which of the following expressions correctly represents the net rate of heat loss by radiation from the sphere?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
 **Primary Concept**: Stefan-Boltzmann Law and Net Radiative Heat Transfer
 **Related Topics**: Thermal Radiation, Emissivity, Blackbody Radiation
 **Prerequisites**: Understanding of the Stefan-Boltzmann Law and the concept of emissivity.
## SOLUTION PATHWAY
**Given Information**:
  Radius of sphere = $r$
  Emissivity of sphere = $e$
  Temperature of sphere = $T$
  Temperature of surroundings = $T_0$
  
**Method & Steps**:
 1. Recall the Stefan-Boltzmann Law, which states that the power radiated by a blackbody is proportional to the fourth power of its absolute temperature.  For a non-blackbody with emissivity $e$, the power radiated is given by:
  $P_{rad} = e \sigma A T^4$, where $A$ is the surface area, $\sigma$ is the Stefan-Boltzmann constant, and $e$ is the emissivity (0 ≤ e ≤ 1).
 2. The sphere's surface area is given by $A = 4\pi r^2$. Substitute this into the equation for power radiated:
  $P_{rad} = e \sigma (4\pi r^2) T^4$
 3.  The sphere also absorbs radiation from the surroundings. The rate of absorption is given by:
  $P_{abs} = e \sigma A T_0^4 = e \sigma (4\pi r^2) T_0^4$
 4. The net rate of heat loss is the difference between the power radiated and the power absorbed:
  $P_{net} = P_{rad} - P_{abs} = e \sigma (4\pi r^2) T^4 - e \sigma (4\pi r^2) T_0^4 = 4\pi r^2 e \sigma (T^4 - T_0^4)$
 5. This is the net rate of heat loss by radiation from the sphere.
  Result: $4\pi r^2 e \sigma (T^4 - T_0^4)$
  Verification: The units are Watts (power), which is consistent with the rate of heat loss.
 ## OPTION ANALYSIS
A :x: Incorrect; This equation lacks the emissivity term 'e'.
B :x: Incorrect; This equation only considers radiation emitted by the sphere, ignoring radiation absorbed from the surroundings.
C :x: Incorrect; This equation uses the volume of the sphere instead of the surface area, and is dimensionally inconsistent for power.
D ✓: Correct; This equation accurately represents the net rate of heat loss by radiation, considering both emission and absorption, incorporating the emissivity and the temperature difference to the fourth power.
 ## LEARNING AIDS
 **Common Mistakes**:
  1. Forgetting the emissivity term 'e':  Remember that real objects don't radiate as perfectly as blackbodies.  The emissivity accounts for this.
  2. Ignoring the absorption of radiation from the surroundings: The net heat loss is the difference between emission and absorption.
 **Quick Tips**:
  1. Always remember the Stefan-Boltzmann Law and its application to non-blackbodies.
  2. Remember that net heat loss = power radiated - power absorbed.

ABDUL JILANI BASHA · Answer

$4\pi r^2 \sigma(T^4 - T_0^4)$

ABDUL JILANI BASHA · Answer

$4\pi r^2 e\sigma T^4$

ABDUL JILANI BASHA · Answer

$\frac{4}{3}\pi r^3 e\sigma(T^4 - T_0^4)$

ABDUL JILANI BASHA · Answer

The correct Option is "$4\pi r^2 e\sigma(T^4 - T_0^4)$"

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