Two identical solid bodies one of aluminium and other of copper are heated to the same temperature and are put in same surrounding. If the emissivity of the aluminium body is 4 times that of copper body, find the ratio of the thermal power radiated by the two bodies. If specific heat of aluminum is `900//kg^(@)C` and that of copper is `390 J//kg^(@)C` and density of copper is 3.4 times that of aluminum, find the ratio of rate of cooling of the two spheres.

Answer

To solve the problem, we will break it down into two parts:

1. Finding the ratio of the thermal power radiated by the two bodies (aluminum and copper).
2. Finding the ratio of the rate of cooling of the two spheres.

### Part 1: Ratio of Thermal Power Radiated

**Step 1: Understand the formula for thermal power radiated.**
The thermal power \( P \) radiated by a body is given by the formula:
\[ P = \varepsilon \sigma A (T^4 - T_0^4) \]
where:
- \( \varepsilon \) = emissivity of the body,
- \( \sigma \) = Stefan-Boltzmann constant,
- \( A \) = surface area of the body,
- \( T \) = absolute temperature of the body,
- \( T_0 \) = absolute temperature of the surroundings.

**Step 2: Set up the ratio of thermal power for aluminum and copper.**
Since both bodies are identical and heated to the same temperature, the area \( A \) and the temperature difference \( (T^4 - T_0^4) \) will be the same for both. Thus, the ratio of thermal power can be simplified to:
\[ \frac{P_{Al}}{P_{Cu}} = \frac{\varepsilon_{Al}}{\varepsilon_{Cu}} \]

**Step 3: Substitute the emissivity values.**
Given that the emissivity of aluminum is 4 times that of copper:
\[ \varepsilon_{Al} = 4 \varepsilon_{Cu} \]
Substituting this into the ratio gives:
\[ \frac{P_{Al}}{P_{Cu}} = \frac{4 \varepsilon_{Cu}}{\varepsilon_{Cu}} = 4 \]

**Final Result for Part 1:**
The ratio of the thermal power radiated by aluminum to copper is:
\[ \frac{P_{Al}}{P_{Cu}} = 4:1 \]

### Part 2: Ratio of Rate of Cooling

**Step 1: Understand the formula for the rate of cooling.**
The rate of cooling \( \frac{dT}{dt} \) is given by:
\[ \frac{dT}{dt} = -\varepsilon \sigma A (T^4 - T_0^4) \]
Similar to thermal power, we can set up the ratio for aluminum and copper:
\[ \frac{\frac{dT_{Al}}{dt}}{\frac{dT_{Cu}}{dt}} = \frac{\varepsilon_{Al} \cdot \rho_{Al} \cdot c_{Al}}{\varepsilon_{Cu} \cdot \rho_{Cu} \cdot c_{Cu}} \]

**Step 2: Substitute the known values.**
Given:
- \( \varepsilon_{Al} = 4 \varepsilon_{Cu} \)
- Density of copper \( \rho_{Cu} = 3.4 \rho_{Al} \)

Substituting these into the ratio gives:
\[ \frac{\frac{dT_{Al}}{dt}}{\frac{dT_{Cu}}{dt}} = \frac{4 \varepsilon_{Cu} \cdot \rho_{Al} \cdot c_{Al}}{\varepsilon_{Cu} \cdot (3.4 \rho_{Al}) \cdot c_{Cu}} \]

**Step 3: Simplify the expression.**
The \( \varepsilon_{Cu} \) and \( \rho_{Al} \) cancel out:
\[ \frac{\frac{dT_{Al}}{dt}}{\frac{dT_{Cu}}{dt}} = \frac{4 \cdot c_{Al}}{3.4 \cdot c_{Cu}} \]

**Step 4: Substitute the specific heat values.**
Given:
- \( c_{Al} = 900 \, \text{J/kg°C} \)
- \( c_{Cu} = 390 \, \text{J/kg°C} \)

Substituting these values gives:
\[ \frac{\frac{dT_{Al}}{dt}}{\frac{dT_{Cu}}{dt}} = \frac{4 \cdot 900}{3.4 \cdot 390} \]

**Step 5: Calculate the ratio.**
Calculating the above expression:
\[ = \frac{3600}{1326} \approx 2.71 \]

**Final Result for Part 2:**
The ratio of the rate of cooling of aluminum to copper is approximately:
\[ \frac{dT_{Al}/dt}{dT_{Cu}/dt} \approx 2.71:1 \]

### Summary of Results:
1. The ratio of thermal power radiated by aluminum to copper is \( 4:1 \).
2. The ratio of the rate of cooling of aluminum to copper is approximately \( 2.71:1 \).