To solve this problem, we need to analyze the heat transfer between the two spherical bodies, A and B, which are in thermal contact but separated by a vacuum. Since both surfaces behave as blackbodies, we can apply the Stefan-Boltzmann law to determine the rate of heat transfer. Let's break this down step by step.

Understanding the System

We have two spherical objects: a solid sphere A with a surface area of 20 m² and a hollow spherical shell B. The initial temperatures are 100°C for A and 20°C for B. The heat capacities are given as 42 J/°C for A and 82 J/°C for B. The thermal conductivity of A is high, while that of B is poor, which means A will lose heat quickly, and B will absorb it slowly.

Applying the Stefan-Boltzmann Law

The rate of heat transfer \( Q \) between two blackbody surfaces can be expressed using the Stefan-Boltzmann law:

\( Q = \sigma \cdot A \cdot (T_A^4 - T_B^4) \)

Where:

• \( \sigma \) is the Stefan-Boltzmann constant (approximately \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \))
• \( A \) is the surface area of the emitting body (in this case, we will consider the area of A since it is emitting heat)
• \( T_A \) and \( T_B \) are the absolute temperatures of A and B in Kelvin.

Converting Temperatures to Kelvin

First, we need to convert the temperatures from Celsius to Kelvin:

• \( T_A = 100 + 273.15 = 373.15 \, \text{K} \)
• \( T_B = 20 + 273.15 = 293.15 \, \text{K} \)

Calculating the Heat Transfer Rate

Now, we can calculate the rate of heat transfer:

\( Q = 5.67 \times 10^{-8} \cdot 20 \cdot (373.15^4 - 293.15^4) \)

Calculating \( T_A^4 \) and \( T_B^4 \):

• \( T_A^4 \approx 1.935 \times 10^{10} \, \text{K}^4 \)
• \( T_B^4 \approx 7.747 \times 10^{9} \, \text{K}^4 \)

Now substituting these values into the equation:

\( Q \approx 5.67 \times 10^{-8} \cdot 20 \cdot (1.935 \times 10^{10} - 7.747 \times 10^{9}) \)

Calculating the difference:

\( Q \approx 5.67 \times 10^{-8} \cdot 20 \cdot 1.160 \times 10^{10} \)

Now, calculating \( Q \):

\( Q \approx 1.32 \, \text{W} \)

Finding the Rate of Change of Temperature

Next, we need to find the rate of change of temperature for both A and B. We can use the formula:

\( \frac{dT}{dt} = \frac{Q}{C} \)

Where \( C \) is the heat capacity of the object.

For Sphere A:

\( \frac{dT_A}{dt} = \frac{-Q}{C_A} = \frac{-1.32}{42} \approx -0.0314 \, \text{°C/s} \)

For Sphere B:

\( \frac{dT_B}{dt} = \frac{Q}{C_B} = \frac{1.32}{82} \approx 0.0161 \, \text{°C/s} \)

Summary of Results

The rate of change of temperature for sphere A is approximately -0.0314 °C/s, indicating it is cooling down, while the rate of change of temperature for sphere B is approximately 0.0161 °C/s, indicating it is warming up. This demonstrates the heat transfer dynamics between the two bodies, with A losing heat and B gaining it.