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 \(\frac{320}{3}\,W\).

To find the temperature at the interface of the rectangular slab, we use the concept of  thermal conduction through two materials in series.

Given:
- Thermal conductivity of iron, \(K_{\text{iron}} = 0.2 \, \text{W/mK}\)
- Thermal conductivity of brass, \(K_{\text{brass}} = 0.3 \, \text{W/mK}\)
- Temperature at one side of the iron slab, \(T_{\text{iron}} = 100^\circ C\)
- Temperature at one side of the brass slab, \(T_{\text{brass}} = 0^\circ C\)
- Thicknesses of both slabs are equal.

Since the slabs are in series and have equal thicknesses, the **heat flux** through both materials is the same. Using the formula for heat conduction, the temperature at the interface \(T_{\text{interface}}\) can be calculated using the ratio of thermal conductivities:

\[
\frac{T_{\text{iron}} - T_{\text{interface}}}{T_{\text{interface}} - T_{\text{brass}}} = \frac{K_{\text{brass}}}{K_{\text{iron}}}
\]

Substitute the known values:

\[
\frac{100 - T_{\text{interface}}}{T_{\text{interface}} - 0} = \frac{0.3}{0.2} = 1.5
\]

Now solve for \(T_{\text{interface}}\):

\[
100 - T_{\text{interface}} = 1.5 T_{\text{interface}}
\]

\[
100 = 2.5 T_{\text{interface}}
\]

\[
T_{\text{interface}} = \frac{100}{2.5} = 40^\circ C
\]

Thus, the temperature at the interface is \(40^\circ C\).

To find the temperature difference (ΔT) between the two ends of the conductor, we can use the formula for heat conduction:

\[
Q = \frac{K \cdot A \cdot \Delta T}{L}
\]

Where:
- \( Q = 6000 \, \text{J/s} \)
- \( K = 200 \, \text{J} \, \text{m}^{-1} \, \text{K}^{-1} \)
- \( A = 0.75 \, \text{m}^2 \)
- \( L = 1 \, \text{m} \)

Rearranging to solve for ΔT:

\[
\Delta T = \frac{Q \cdot L}{K \cdot A}
\]

\[
\Delta T = \frac{6000 \cdot 1}{200 \cdot 0.75} = 40 \, \text{K}
\]

The temperature difference between the two ends is 40 K or 40 °C

Given that the thermal resistance is the same for both rods, we have the equation:

\[
\frac{L_1}{K_1} = \frac{L_2}{K_2}
\]

So, the ratio of lengths is:

\[
\frac{L_1}{L_2} = \frac{K_1}{K_2}
\]

If the ratio of the thermal conductivities \(K_1:K_2 = 5:4\), the ratio of the lengths will be the same:

\[
\frac{L_1}{L_2} = \frac{5}{4}
\]

Therefore, the correct ratio of the lengths is \(5:4\).

1. Heat released by \(x\) gm steam at \(100^\circ C\):
\[
Q_{\text{steam}} = x \times 540
\]

2. Heat required to convert \(y\) gm ice at \(0^\circ C\) to water at \(100^\circ C\):
\[
Q_{\text{ice}} = y \times 80 + y \times 100 = y \times 180
\]

3. Equating:
\[
x \times 540 = y \times 180
\]

\[
\frac{y}{x} = 3
\]

So, \( y : x = 3 : 1 \).

All the given statements are correct:

• A. Work and heat depend on the path taken during a process, so they are path functions.
• B. Internal energy depends only on the state of the system, making it a state function.
• C. For gases,
  $C_P > C_V$ 

  because extra heat is required at constant pressure to do expansion work.

• D. At constant volume, the gas does no work since volume does not change (
  $W = P\Delta V = 0$ 

  ).