Calcualte the maximum amount of heat which may be lost per second by radiation from a sphere of 10 cm in diameter and at a temperature of `227^(@)`C when placed in an encloser at a temperature of `27^(@)`C. `sigma = 5.7xx10^(-8)Wm^(-2)K^(-4)`.

Explanation

To solve the problem of calculating the maximum amount of heat lost per second by radiation from a sphere, we will follow these steps:

### Step 1: Convert the temperatures from Celsius to Kelvin
- The temperature of the sphere \( T \) is given as \( 227^\circ C \).
- To convert to Kelvin:
\[ T = 227 + 273 = 500 \, K \]
- The temperature of the surroundings \( T_0 \) is given as \( 27^\circ C \).
- To convert to Kelvin:
\[ T_0 = 27 + 273 = 300 \, K \]

### Step 2: Calculate the radius of the sphere
- The diameter of the sphere is \( 10 \, cm \), so the radius \( r \) is:
\[ r = \frac{10 \, cm}{2} = 5 \, cm = 0.05 \, m \]

### Step 3: Calculate the surface area of the sphere
- The surface area \( A \) of a sphere is given by the formula:
\[ A = 4 \pi r^2 \]
- Substituting the radius:
\[ A = 4 \pi (0.05)^2 = 4 \pi (0.0025) = 0.0314 \, m^2 \]

### Step 4: Use the Stefan-Boltzmann Law to calculate the heat loss
- The formula for the power radiated by a black body is:
\[ P = \sigma A (T^4 - T_0^4) \]
- Where \( \sigma = 5.7 \times 10^{-8} \, W/m^2 K^4 \).
- Substituting the values:
\[ P = 5.7 \times 10^{-8} \times 0.0314 \times (500^4 - 300^4) \]

### Step 5: Calculate \( T^4 \) and \( T_0^4 \)
- Calculate \( 500^4 \):
\[ 500^4 = 62500000000 \]
- Calculate \( 300^4 \):
\[ 300^4 = 8100000000 \]
- Now, subtract these values:
\[ 500^4 - 300^4 = 62500000000 - 8100000000 = 54400000000 \]

### Step 6: Substitute back to find \( P \)
- Now substituting back into the power equation:
\[ P = 5.7 \times 10^{-8} \times 0.0314 \times 54400000000 \]
- Calculate:
\[ P = 5.7 \times 0.0314 \times 54400000000 \approx 97.36 \, W \]

### Step 7: Convert Joules per second to Calories per second
- Since \( 1 \, J = 0.239 \, cal \):
\[ \text{Calories per second} = \frac{97.36}{4.2} \approx 23.18 \, cal/s \]

### Final Answer
The maximum amount of heat lost per second by radiation from the sphere is approximately **97.36 Joules per second** or **23.18 Calories per second**.

---