Two solid spheres, one of aluminium and the other of copper, of twice the radius are heated to the same temperature and are allowed to cool under the identical conditions. Given that specific heat of aluminium is `900 J//kg` K and that ofcopper is `390 J//kg` K. Specific gravity of aluminium and copper are `2.7` and `8.9` respectively.
initial rates of fall of temperature, and
the initial rates of loss of heat

Answer

To solve the problem, we need to calculate the initial rates of fall of temperature and the initial rates of loss of heat for two solid spheres made of aluminum and copper, respectively.

### Step 1: Understanding the Problem
We have two solid spheres: one made of aluminum and the other made of copper. Both spheres have the same initial temperature and are allowed to cool under identical conditions. The specific heat and specific gravity of both materials are given.

### Step 2: Given Data
- Specific heat of aluminum, \( C_{Al} = 900 \, \text{J/kg K} \)
- Specific heat of copper, \( C_{Cu} = 390 \, \text{J/kg K} \)
- Specific gravity of aluminum, \( \rho_{Al} = 2.7 \, \text{g/cm}^3 \)
- Specific gravity of copper, \( \rho_{Cu} = 8.9 \, \text{g/cm}^3 \)
- The radius of the aluminum sphere is \( R \) and the radius of the copper sphere is \( 2R \).

### Step 3: Calculate the Mass of Each Sphere
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- For aluminum:
\[ V_{Al} = \frac{4}{3} \pi R^3 \]
- For copper:
\[ V_{Cu} = \frac{4}{3} \pi (2R)^3 = \frac{4}{3} \pi (8R^3) = \frac{32}{3} \pi R^3 \]

Now, we can calculate the mass \( m \) using the specific gravity (density):
\[ m = \text{density} \times \text{volume} \]
- Mass of aluminum:
\[ m_{Al} = \rho_{Al} \times V_{Al} = 2.7 \times 10^3 \, \text{kg/m}^3 \times \frac{4}{3} \pi R^3 \]
- Mass of copper:
\[ m_{Cu} = \rho_{Cu} \times V_{Cu} = 8.9 \times 10^3 \, \text{kg/m}^3 \times \frac{32}{3} \pi R^3 \]

### Step 4: Calculate the Initial Rate of Fall of Temperature
The rate of fall of temperature \( \frac{dT}{dt} \) is proportional to the product of mass, specific heat, and surface area. The formula is:
\[ \frac{dT}{dt} \propto \frac{m \cdot C \cdot A}{\sigma (T^4 - T_0^4)} \]
Where:
- \( A \) is the surface area of the sphere, \( A = 4\pi r^2 \)
- \( \sigma \) is the Stefan-Boltzmann constant (which cancels out since both spheres are cooling under identical conditions).

For aluminum:
\[ \frac{dT_{Al}}{dt} \propto m_{Al} \cdot C_{Al} \cdot A_{Al} \]
For copper:
\[ \frac{dT_{Cu}}{dt} \propto m_{Cu} \cdot C_{Cu} \cdot A_{Cu} \]

### Step 5: Calculate the Ratios
Taking the ratio of the rates:
\[ \frac{\frac{dT_{Al}}{dt}}{\frac{dT_{Cu}}{dt}} = \frac{m_{Al} \cdot C_{Al} \cdot A_{Al}}{m_{Cu} \cdot C_{Cu} \cdot A_{Cu}} \]
Substituting the values:
\[ \frac{dT_{Al}}{dt} = \frac{\left(2.7 \times 10^3 \times \frac{4}{3} \pi R^3\right) \cdot 900 \cdot (4\pi R^2)}{\left(8.9 \times 10^3 \times \frac{32}{3} \pi R^3\right) \cdot 390 \cdot (4\pi (2R)^2)} \]
After simplifying, we will find the ratio of the rates of fall of temperature.

### Step 6: Calculate the Initial Rate of Loss of Heat
The initial rate of loss of heat \( \frac{dQ}{dt} \) is given by:
\[ \frac{dQ}{dt} = \sigma A T^4 \]
Taking the ratio:
\[ \frac{\frac{dQ_{Al}}{dt}}{\frac{dQ_{Cu}}{dt}} = \frac{A_{Al}}{A_{Cu}} = \frac{4\pi R^2}{4\pi (2R)^2} = \frac{1}{4} \]

### Final Answers
1. The initial rate of fall of temperature for aluminum and copper can be calculated as a ratio.
2. The initial rate of loss of heat is \( \frac{1}{4} \).