Solution:

To solve this, we use Wien's Displacement Law, which states:

\[
\lambda_{\text{max}} = \frac{b}{T}
\]

Where:
- \(\lambda_{\text{max}}\) is the wavelength at which the maximum energy is emitted,
- \(b = 2.88 \times 10^6 \, \text{nmK}\) is Wien's constant,
- \(T = 5760 \, K\) is the temperature of the black body.

Step 1: Calculate \(\lambda_{\text{max}}\)
\[
\lambda_{\text{max}} = \frac{2.88 \times 10^6}{5760} = 500 \, \text{nm}
\]

This means that the maximum energy is emitted at a wavelength of 500 nm.

 Step 2: Compare the energies at different wavelengths

- At 500 nm (\( \lambda_{\text{max}} \)): This is where the black body emits the maximum energy. So, \(U_2\) will be the largest.
- At 250 nm: This wavelength is shorter than \(\lambda_{\text{max}}\), and the energy decreases as we move away from the peak, so \(U_1 < U_2\).
- At 1000 nm: This wavelength is longer than \(\lambda_{\text{max}}\), and the energy decreases further, so \(U_3 < U_2\).

Thus, \(U_2 > U_1\), and the correct comparison is \(U_2 > U_1\).

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Solution:

The color change in the heated iron from dull red to reddish yellow to white hot can be explained using Wein's Displacement Law. This law states that the wavelength (\(\lambda_{max}\)) at which the maximum emission occurs is inversely proportional to the absolute temperature \(T\) of the object:

\[
\lambda_{max} \propto \frac{1}{T}
\]

As the iron's temperature increases, the peak wavelength of emitted radiation shifts toward shorter wavelengths (from red to yellow and eventually to white, which is a combination of all visible wavelengths). Thus, the observed color change corresponds to an increase in temperature, with shorter wavelengths being emitted at higher temperatures.

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Solution:

The energy emitted by a black body is proportional to the fourth power of its absolute temperature, as per Stefan-Boltzmann Law:

\[
P \propto T^4
\]

Step 1: Convert temperature to kelvins

The temperature of the black body is \(727^\circ C\), which is:

\[
T = 727 + 273 = 1000\,K
\]

Step 2: Apply Stefan-Boltzmann Law

Since the energy emitted is proportional to \(T^4\):

\[
P \propto (1000)^4
\]

Thus, the rate of energy emission is proportional to \((1000)^4\).

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Solution:

To solve this, we use Stefan-Boltzmann Law for radiative heat transfer:

\[
P = \sigma A (T^4 - T_s^4)
\]

Where:
- \(P\) is the power (rate of energy loss),
- \(\sigma\) is the Stefan-Boltzmann constant,
- \(A\) is the area of the body,
- \(T\) is the temperature of the body in kelvins,
- \(T_s\) is the temperature of the surroundings in kelvins.

 Step 1: Convert temperatures to kelvins
- Initial temperature of the black body: \(727^\circ C = 727 + 273 = 1000\,K\)
- Surrounding temperature: \(227^\circ C = 227 + 273 = 500\,K\)
- New temperature of the black body: \(1227^\circ C = 1227 + 273 = 1500\,K\)

 Step 2: Ratio of power losses
Let the power at \(T_1 = 1000\,K\) be \(P_1 = 20\,W\). The new power at \(T_2 = 1500\,K\) is \(P_2\). Using the Stefan-Boltzmann law:

\[
\frac{P_2}{P_1} = \frac{T_2^4 - T_s^4}{T_1^4 - T_s^4}
\]

Substitute the values:

\[
\frac{P_2}{20} = \frac{1500^4 - 500^4}{1000^4 - 500^4}
\]

Step 3: Simplify the powers of temperatures
\[
\frac{P_2}{20} = \frac{1500^4 - 500^4}{1000^4 - 500^4}
\]

You can approximate the values:

\[
P_2 = 20 \times \frac{(1500^4 - 500^4)}{(1000^4 - 500^4)}
\]

After simplifying, you'll find:

\[
P_2 = \frac{320}{3} \text{ watts}
\]

Thus, the rate of energy loss at 1227°C is \(\frac{320}{3}\,W\).

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