A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identical surrounding temperatures. Assume that the emisssivity of both the spheres is the same. Find ratio of (a) the rate of heat loss from the aluminium sphere to the rate of heat loss from the copper sphere and (b) the rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of copper sphere. The specific heat capacity of aluminium `=900Jkg^(-1)C^(-1)` . and that of copper `=390Jkg^(-1)C^(-1)` . The density of copper `=3.4` times the density of aluminium.

Explanation

To solve the problem step by step, we will break down the calculations for both parts (a) and (b) of the question.

### Given Data:
- Specific heat capacity of Aluminium, \( c_{Al} = 900 \, J \, kg^{-1} \, C^{-1} \)
- Specific heat capacity of Copper, \( c_{Cu} = 390 \, J \, kg^{-1} \, C^{-1} \)
- Density of Copper, \( \rho_{Cu} = 3.4 \times \rho_{Al} \)
- Radius of Copper sphere, \( R_{Cu} = 2 \times R_{Al} \)

### Step 1: Calculate the Mass of Each Sphere
1. **Volume of a sphere**:
\[ V = \frac{4}{3} \pi R^3 \]
2. **Mass of Aluminium sphere**:
\[ m_{Al} = \rho_{Al} \cdot V_{Al} = \rho_{Al} \cdot \frac{4}{3} \pi R_{Al}^3 \]
3. **Mass of Copper sphere**:
\[ m_{Cu} = \rho_{Cu} \cdot V_{Cu} = \rho_{Cu} \cdot \frac{4}{3} \pi (2 R_{Al})^3 = \rho_{Cu} \cdot \frac{4}{3} \pi (8 R_{Al}^3) = 8 \rho_{Cu} \cdot \frac{4}{3} \pi R_{Al}^3 \]
Since \( \rho_{Cu} = 3.4 \times \rho_{Al} \):
\[ m_{Cu} = 8 \cdot 3.4 \cdot \rho_{Al} \cdot \frac{4}{3} \pi R_{Al}^3 = 27.2 \cdot m_{Al} \]

### Step 2: Calculate the Ratio of Masses
The ratio of the mass of Aluminium to the mass of Copper is:
\[ \frac{m_{Al}}{m_{Cu}} = \frac{1}{27.2} \]

### Step 3: Calculate the Rate of Heat Loss
Using the Stefan-Boltzmann law:
\[ \text{Rate of heat loss} = \epsilon \sigma A T^4 \]
Where \( A \) is the surface area, given by:
\[ A = 4 \pi R^2 \]
1. **Area of Aluminium sphere**:
\[ A_{Al} = 4 \pi R_{Al}^2 \]
2. **Area of Copper sphere**:
\[ A_{Cu} = 4 \pi (2 R_{Al})^2 = 16 \pi R_{Al}^2 \]
3. **Rate of heat loss from Aluminium**:
\[ Q_{Al} = \epsilon \sigma A_{Al} T^4 = \epsilon \sigma (4 \pi R_{Al}^2) T^4 \]
4. **Rate of heat loss from Copper**:
\[ Q_{Cu} = \epsilon \sigma A_{Cu} T^4 = \epsilon \sigma (16 \pi R_{Al}^2) T^4 \]

### Step 4: Ratio of Heat Loss
The ratio of the rate of heat loss from Aluminium to Copper is:
\[ \frac{Q_{Al}}{Q_{Cu}} = \frac{4 \pi R_{Al}^2}{16 \pi R_{Al}^2} = \frac{1}{4} \]

### Step 5: Calculate the Rate of Fall of Temperature
Using the formula:
\[ \text{Rate of heat loss} = m c \frac{dT}{dt} \]
1. For Aluminium:
\[ Q_{Al} = m_{Al} c_{Al} \frac{dT_{Al}}{dt} \]
2. For Copper:
\[ Q_{Cu} = m_{Cu} c_{Cu} \frac{dT_{Cu}}{dt} \]

### Step 6: Set Up the Equation
From the heat loss equations:
\[ m_{Al} c_{Al} \frac{dT_{Al}}{dt} = \frac{1}{4} m_{Cu} c_{Cu} \frac{dT_{Cu}}{dt} \]
Substituting \( m_{Cu} = 27.2 m_{Al} \):
\[ m_{Al} c_{Al} \frac{dT_{Al}}{dt} = \frac{1}{4} (27.2 m_{Al}) c_{Cu} \frac{dT_{Cu}}{dt} \]
Cancelling \( m_{Al} \):
\[ c_{Al} \frac{dT_{Al}}{dt} = \frac{27.2}{4} c_{Cu} \frac{dT_{Cu}}{dt} \]

### Step 7: Ratio of Rate of Fall of Temperature
Rearranging gives:
\[ \frac{dT_{Al}}{dT_{Cu}} = \frac{27.2}{4} \cdot \frac{c_{Cu}}{c_{Al}} = \frac{27.2}{4} \cdot \frac{390}{900} \]
Calculating:
\[ \frac{dT_{Al}}{dT_{Cu}} = \frac{27.2 \cdot 390}{4 \cdot 900} = \frac{10608}{3600} \approx 2.94 \]

### Final Answers:
(a) The ratio of the rate of heat loss from the Aluminium sphere to the Copper sphere is \( \frac{1}{4} \).

(b) The ratio of the rate of fall of temperature of the Aluminium sphere to the Copper sphere is approximately \( 2.94 \).