Two spheres of the same materials have radii 1 m and 4 m and temperatures 4000 K and 2000 K resectively the energy radiated per second by the first sphere is

Options

A. greater than that by the second
B. less than that by the second
C. equal in both cases
D. the information is incomplete to draw any conclusion.

Explanation

To solve the problem of finding the energy radiated per second by the first sphere, we can follow these steps:

### Step 1: Understand the given data
We have two spheres:
- Sphere 1:
- Radius \( R_1 = 1 \, \text{m} \)
- Temperature \( T_1 = 4000 \, \text{K} \)

- Sphere 2:
- Radius \( R_2 = 4 \, \text{m} \)
- Temperature \( T_2 = 2000 \, \text{K} \)

### Step 2: Use Stefan-Boltzmann Law
According to the Stefan-Boltzmann Law, the power radiated per second (P) by a black body is given by:
\[ P = \sigma \cdot e \cdot A \cdot T^4 \]
where:
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( e \) is the emissivity (which is the same for both spheres since they are made of the same material),
- \( A \) is the surface area,
- \( T \) is the absolute temperature.

### Step 3: Calculate the surface area of the spheres
The surface area \( A \) of a sphere is given by:
\[ A = 4\pi R^2 \]
Thus, for Sphere 1:
\[ A_1 = 4\pi R_1^2 = 4\pi (1)^2 = 4\pi \, \text{m}^2 \]
And for Sphere 2:
\[ A_2 = 4\pi R_2^2 = 4\pi (4)^2 = 64\pi \, \text{m}^2 \]

### Step 4: Write the power equations for both spheres
For Sphere 1:
\[ P_1 = \sigma \cdot e \cdot A_1 \cdot T_1^4 = \sigma \cdot e \cdot (4\pi) \cdot (4000)^4 \]

For Sphere 2:
\[ P_2 = \sigma \cdot e \cdot A_2 \cdot T_2^4 = \sigma \cdot e \cdot (64\pi) \cdot (2000)^4 \]

### Step 5: Set up the ratio of the powers
To find the relationship between \( P_1 \) and \( P_2 \), we can divide the two equations:
\[ \frac{P_1}{P_2} = \frac{(4\pi) \cdot (4000)^4}{(64\pi) \cdot (2000)^4} \]

### Step 6: Simplify the expression
The \( \pi \) cancels out:
\[ \frac{P_1}{P_2} = \frac{4 \cdot (4000)^4}{64 \cdot (2000)^4} = \frac{4}{64} \cdot \left(\frac{4000}{2000}\right)^4 \]
\[ = \frac{1}{16} \cdot (2)^4 = \frac{1}{16} \cdot 16 = 1 \]

### Step 7: Conclude the result
Since \( \frac{P_1}{P_2} = 1 \), it follows that:
\[ P_1 = P_2 \]
Thus, the energy radiated per second by the first sphere is equal to that of the second sphere.

### Final Answer
The energy radiated per second by the first sphere is equal to the energy radiated per second by the second sphere.

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