Two bodies A and B are placed in an evacuated vessel maintained at a temperature of `27^(@)C`, the temperature of A is `327^(@)C` and that of B is `227^(@)C`. Then the ratio of heat loss by body A and B is

Options

A. `3:2`
B. `245:1`
C. `2:1`
D. `3:1`

Explanation

To solve the problem, we need to find the ratio of heat loss by bodies A and B using Newton's Law of Cooling.

### Step-by-Step Solution:

1. **Identify the temperatures**:
- Temperature of body A, \( T_A = 327^\circ C \)
- Temperature of body B, \( T_B = 227^\circ C \)
- Surrounding temperature, \( T_s = 27^\circ C \)

2. **Convert temperatures to Kelvin**:
- To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
- For body A:
\[ T_A = 327 + 273 = 600 \, K \]
- For body B:
\[ T_B = 227 + 273 = 500 \, K \]
- For surrounding temperature:
\[ T_s = 27 + 273 = 300 \, K \]

3. **Apply Newton's Law of Cooling**:
- According to Newton's Law of Cooling, the rate of heat loss \( \frac{dQ}{dt} \) is given by:
\[ \frac{dQ}{dt} = k (T - T_s) \]
- For body A:
\[ \frac{dQ_A}{dt} = k (T_A - T_s) = k (600 - 300) = k \cdot 300 \]
- For body B:
\[ \frac{dQ_B}{dt} = k (T_B - T_s) = k (500 - 300) = k \cdot 200 \]

4. **Calculate the ratio of heat loss**:
- Now, we can find the ratio of heat loss by A and B:
\[ \frac{dQ_A/dt}{dQ_B/dt} = \frac{k \cdot 300}{k \cdot 200} = \frac{300}{200} = \frac{3}{2} \]

5. **Conclusion**:
- Therefore, the ratio of heat loss by body A to body B is \( \frac{3}{2} \).

### Final Answer:
The ratio of heat loss by body A and B is \( \frac{3}{2} \).

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