Compare the rate of radiation of metal body at 727 `""^(@)C` and 227 `""^(@)C`.

Answer

To compare the rate of radiation of a metal body at two different temperatures, we can use the Stefan-Boltzmann Law, which states that the power radiated per unit area of a black body is directly proportional to the fourth power of its absolute temperature (in Kelvin). The formula is given by:

\[ \frac{dQ}{dt} = E \cdot A \cdot \sigma \cdot T^4 \]

Where:
- \(dQ/dt\) is the rate of radiation,
- \(E\) is the emissivity (which we can assume to be constant for both temperatures),
- \(A\) is the area (also constant for both),
- \(\sigma\) is the Stefan-Boltzmann constant (constant for both),
- \(T\) is the absolute temperature in Kelvin.

### Step 1: Convert temperatures from Celsius to Kelvin
- For \(T_1 = 727^\circ C\):
\[ T_1 = 727 + 273 = 1000 \, K \]
- For \(T_2 = 227^\circ C\):
\[ T_2 = 227 + 273 = 500 \, K \]

### Step 2: Write the expressions for the rates of radiation
Using the formula for both temperatures:
\[ \frac{dQ}{dt}\bigg|_{T_1} = E \cdot A \cdot \sigma \cdot (1000)^4 \]
\[ \frac{dQ}{dt}\bigg|_{T_2} = E \cdot A \cdot \sigma \cdot (500)^4 \]

### Step 3: Set up the ratio of the rates of radiation
Now, we can find the ratio of the rates of radiation at the two temperatures:
\[ \frac{\frac{dQ}{dt}\bigg|_{T_1}}{\frac{dQ}{dt}\bigg|_{T_2}} = \frac{E \cdot A \cdot \sigma \cdot (1000)^4}{E \cdot A \cdot \sigma \cdot (500)^4} \]

### Step 4: Simplify the ratio
The constants \(E\), \(A\), and \(\sigma\) cancel out:
\[ \frac{dQ}{dt}\bigg|_{T_1} : \frac{dQ}{dt}\bigg|_{T_2} = \frac{(1000)^4}{(500)^4} = \left(\frac{1000}{500}\right)^4 = 2^4 = 16 \]

### Conclusion
The rate of radiation of the metal body at \(727^\circ C\) is 16 times greater than that at \(227^\circ C\).

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