A ball is coated with lamp black. Its temperature is `327^(@)C` and is placed in the atmosphere at `27^(@)C`. Let the rate of cooling be R. If the temperature of the ball be `627^(@)C`, what will be its rate of cooling ?

Options

A. 2 R
B. 4 R
C. 8 R
D. `(16)/(3) R`

Explanation

To solve the problem, we will use the concept of Newton's Law of Cooling, which states that the rate of cooling of an object is proportional to the difference in temperature between the object and its surroundings. In this case, we will also consider the Stefan-Boltzmann Law for radiative heat transfer, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- The initial temperature of the ball is \(327^\circ C\).
- Convert to Kelvin:
\[ T_1 = 327 + 273 = 600 \, K \]
- The surrounding temperature is \(27^\circ C\):
- Convert to Kelvin:
\[ T_s = 27 + 273 = 300 \, K \]

2. **Calculate the Rate of Cooling (R1) at Initial Temperature:**
- According to the Stefan-Boltzmann Law, the rate of cooling \(R_1\) can be expressed as:
\[ R_1 \propto T_1^4 - T_s^4 \]
- Substitute the values:
\[ R_1 \propto 600^4 - 300^4 \]

3. **Calculate the New Temperature of the Ball:**
- The new temperature of the ball is \(627^\circ C\).
- Convert to Kelvin:
\[ T_2 = 627 + 273 = 900 \, K \]

4. **Calculate the Rate of Cooling (R2) at New Temperature:**
- The rate of cooling \(R_2\) can be expressed as:
\[ R_2 \propto T_2^4 - T_s^4 \]
- Substitute the values:
\[ R_2 \propto 900^4 - 300^4 \]

5. **Set Up the Ratio of Rates of Cooling:**
- The ratio of the rates of cooling \( \frac{R_1}{R_2} \) can be expressed as:
\[ \frac{R_1}{R_2} = \frac{T_1^4 - T_s^4}{T_2^4 - T_s^4} \]

6. **Calculate the Values:**
- Calculate \(600^4\) and \(300^4\):
\[ 600^4 = 1.296 \times 10^{11}, \quad 300^4 = 8.1 \times 10^{9} \]
- Therefore,
\[ R_1 \propto 1.296 \times 10^{11} - 8.1 \times 10^{9} \approx 1.215 \times 10^{11} \]
- Calculate \(900^4\):
\[ 900^4 = 6.561 \times 10^{11} \]
- Therefore,
\[ R_2 \propto 6.561 \times 10^{11} - 8.1 \times 10^{9} \approx 6.48 \times 10^{11} \]

7. **Final Calculation of the Ratio:**
- Now, calculate the ratio:
\[ \frac{R_1}{R_2} = \frac{1.215 \times 10^{11}}{6.48 \times 10^{11}} \approx \frac{3}{16} \]
- Therefore, we find:
\[ R_2 = \frac{16}{3} R_1 \]

### Conclusion:
The rate of cooling when the temperature of the ball is \(627^\circ C\) will be \( \frac{16}{3} R \), where \(R\) is the initial rate of cooling.