Two solid spheres of radii `R_(1)` and `R_(2)` are made of the same material and have similar surfaces. These are raised to the same temperature and then allowed to cool under identical conditions. The ratio of their initial rates of loss of heat are

Options

A. `R_1/R_2`
B. `R_2/R_1`
C. `R_1^2/R_2^2`
D. `R_2^2/R_1^2`

Answer

To solve the problem of finding the ratio of the initial rates of loss of heat for two solid spheres with radii \( R_1 \) and \( R_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Heat Loss by Radiation**:
The rate of heat loss \( \dot{Q} \) from a body due to radiation can be expressed using the Stefan-Boltzmann law:
\[ \dot{Q} = e \sigma A (T^4 - T_s^4) \]
where:
- \( e \) is the emissivity of the surface (same for both spheres),
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the sphere,
- \( T \) is the temperature of the sphere,
- \( T_s \) is the surrounding temperature.

2. **Calculating Surface Area**:
The surface area \( A \) of a solid sphere is given by:
\[ A = 4 \pi R^2 \]
Therefore, for the two spheres:
- For sphere 1: \( A_1 = 4 \pi R_1^2 \)
- For sphere 2: \( A_2 = 4 \pi R_2^2 \)

3. **Substituting Surface Area into Heat Loss Equation**:
Now substituting the surface area into the heat loss equation for both spheres:
\[ \dot{Q_1} = e \sigma (4 \pi R_1^2) (T^4 - T_s^4) \]
\[ \dot{Q_2} = e \sigma (4 \pi R_2^2) (T^4 - T_s^4) \]

4. **Finding the Ratio of Heat Loss**:
To find the ratio of the initial rates of loss of heat:
\[ \frac{\dot{Q_1}}{\dot{Q_2}} = \frac{e \sigma (4 \pi R_1^2) (T^4 - T_s^4)}{e \sigma (4 \pi R_2^2) (T^4 - T_s^4)} \]
The constants \( e \), \( \sigma \), \( 4 \pi \), and \( (T^4 - T_s^4) \) cancel out:
\[ \frac{\dot{Q_1}}{\dot{Q_2}} = \frac{R_1^2}{R_2^2} \]

5. **Final Result**:
Thus, the ratio of their initial rates of loss of heat is:
\[ \frac{\dot{Q_1}}{\dot{Q_2}} = \frac{R_1^2}{R_2^2} \]