If a spherical blackbody’s temperature is increased from T to 2T, while its volume is reduced to 1/8 times of its initial value, then the ratio of new radiant power to the initial radiant power is x. Then the value of x is______.

Answer

To solve the problem, we need to determine the ratio of the new radiant power (P2) to the initial radiant power (P1) of a spherical black body when its temperature is increased and its volume is reduced.

### Step-by-Step Solution:

1. **Understanding Radiant Power**:
The radiant power (P) emitted by a black body is given by Stefan-Boltzmann Law:
\[ P = \sigma A T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the absolute temperature.

2. **Initial Conditions**:
Let the initial temperature be \( T \) and the initial radius be \( r \). The initial volume \( V_1 \) of the sphere is:
\[ V_1 = \frac{4}{3} \pi r^3 \]
The initial surface area \( A_1 \) is:
\[ A_1 = 4 \pi r^2 \]
Thus, the initial radiant power \( P_1 \) is:
\[ P_1 = \sigma A_1 T^4 = \sigma (4 \pi r^2) T^4 \]

3. **New Conditions**:
The temperature is increased to \( 2T \) and the volume is reduced to \( \frac{1}{8} \) of the initial volume. Therefore, the new volume \( V_2 \) is:
\[ V_2 = \frac{1}{8} V_1 = \frac{1}{8} \left( \frac{4}{3} \pi r^3 \right) = \frac{1}{6} \pi r^3 \]
Let the new radius be \( r' \). Then:
\[ V_2 = \frac{4}{3} \pi (r')^3 \]
Setting the two expressions for volume equal gives:
\[ \frac{4}{3} \pi (r')^3 = \frac{1}{8} \left( \frac{4}{3} \pi r^3 \right) \]
Simplifying, we find:
\[ (r')^3 = \frac{r^3}{8} \implies r' = \frac{r}{2} \]

4. **New Surface Area**:
The new surface area \( A_2 \) is:
\[ A_2 = 4 \pi (r')^2 = 4 \pi \left( \frac{r}{2} \right)^2 = 4 \pi \left( \frac{r^2}{4} \right) = \pi r^2 \]

5. **New Radiant Power**:
The new radiant power \( P_2 \) is:
\[ P_2 = \sigma A_2 (2T)^4 = \sigma (\pi r^2) (16 T^4) = 16 \sigma \pi r^2 T^4 \]

6. **Ratio of New Radiant Power to Initial Radiant Power**:
Now we can find the ratio \( \frac{P_2}{P_1} \):
\[ \frac{P_2}{P_1} = \frac{16 \sigma \pi r^2 T^4}{\sigma (4 \pi r^2) T^4} \]
Simplifying this gives:
\[ \frac{P_2}{P_1} = \frac{16}{4} = 4 \]

### Final Answer:
The value of \( x \) (the ratio of new radiant power to the initial radiant power) is:
\[ \boxed{4} \]