If `T_(B)` and `T_(s)` are the temperatures of the body and the surroundings and `T_(B) -T_(s)` is of very high value, then the rate of cooling in natural convection is proportional to .

Options

A. `T_(B)^(4)`
B. `T_(S)^(4)`
C. `(T_(B)-T_(S))^(5/4)`
D. `(T_(B)^(5/4)-T_(S)^(5/4))`

Answer

To solve the problem, we need to determine how the rate of cooling in natural convection relates to the temperature difference between a body and its surroundings when that temperature difference is very high.

### Step-by-step Solution:

1. **Understanding the Variables**:
- Let \( T_B \) be the temperature of the body.
- Let \( T_s \) be the temperature of the surroundings.
- The difference \( T_B - T_s \) is given to be very high.

2. **Applying Newton's Law of Cooling**:
- In natural convection, the rate of cooling is generally described by Newton's Law of Cooling, which states that the rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings.
- Mathematically, this can be expressed as:
\[ \frac{dQ}{dt} \propto (T_B - T_s) \]
- Here, \( \frac{dQ}{dt} \) is the rate of heat loss.

3. **Considering High Temperature Difference**:
- Since \( T_B - T_s \) is very high, we can say that the body is significantly hotter than the surroundings.
- This implies that the temperature of the surroundings \( T_s \) can be considered negligible compared to \( T_B \).

4. **Simplifying the Expression**:
- If we denote \( \Delta T = T_B - T_s \), then for very high values of \( \Delta T \):
\[ \frac{dQ}{dt} \propto \Delta T \]
- This means that the rate of cooling is directly proportional to the temperature difference \( \Delta T \).

5. **Conclusion**:
- Therefore, the rate of cooling in natural convection is proportional to the temperature difference \( T_B - T_s \) when this difference is very high.

### Final Answer:
The rate of cooling in natural convection is proportional to \( T_B - T_s \).

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