Two uniform solid spheres made of copper have radii 15 cm and 20 cm respectively. Both of them are heated to a temperature of `70^(@)C` and then placed in a region of ambient temperature equal to `45^(@)C` . What will be the ratio of the initial rates of cooling of both the spheres?

Answer

To solve the problem, we need to determine the ratio of the initial rates of cooling of two solid copper spheres with different radii when placed in an ambient temperature. We will use Newton's Law of Cooling, which states that the rate of heat loss of a body is proportional to the difference in temperature between the body and its surroundings.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Radius of Sphere 1, \( R_1 = 15 \, \text{cm} \)
- Radius of Sphere 2, \( R_2 = 20 \, \text{cm} \)
- Initial temperature of both spheres, \( T_i = 70^\circ C \)
- Ambient temperature, \( T_a = 45^\circ C \)

2. **Calculate the Temperature Difference:**
- The temperature difference for both spheres is the same:
\[ \Delta T = T_i - T_a = 70^\circ C - 45^\circ C = 25^\circ C \]

3. **Determine the Surface Area of Each Sphere:**
- The surface area \( A \) of a sphere is given by the formula:
\[ A = 4\pi R^2 \]
- For Sphere 1:
\[ A_1 = 4\pi (15)^2 = 4\pi \times 225 = 900\pi \, \text{cm}^2 \]
- For Sphere 2:
\[ A_2 = 4\pi (20)^2 = 4\pi \times 400 = 1600\pi \, \text{cm}^2 \]

4. **Calculate the Mass of Each Sphere:**
- The mass \( m \) of a sphere is given by:
\[ m = \frac{4}{3}\pi R^3 \rho \]
- Assuming the density \( \rho \) of copper is constant, we can express the mass in terms of volume:
- For Sphere 1:
\[ m_1 = \frac{4}{3}\pi (15)^3 \rho = \frac{4}{3}\pi \times 3375 \rho = 4500\pi \rho \, \text{g} \]
- For Sphere 2:
\[ m_2 = \frac{4}{3}\pi (20)^3 \rho = \frac{4}{3}\pi \times 8000 \rho = 10667\pi \rho \, \text{g} \]

5. **Apply Newton's Law of Cooling:**
- The rate of cooling \( \frac{dq}{dt} \) is proportional to the surface area and the temperature difference:
\[ \frac{dq}{dt} \propto A \Delta T \]
- Therefore, the initial rates of cooling for both spheres can be expressed as:
\[ \frac{h_1}{h_2} = \frac{A_1}{A_2} \]

6. **Calculate the Ratio of the Initial Rates of Cooling:**
- Substitute the surface areas:
\[ \frac{h_1}{h_2} = \frac{900\pi}{1600\pi} = \frac{900}{1600} = \frac{9}{16} \]

7. **Final Ratio of Initial Rates of Cooling:**
- The ratio of the initial rates of cooling of the two spheres is:
\[ \frac{h_1}{h_2} = \frac{9}{16} \]

### Summary:
The ratio of the initial rates of cooling of the two spheres is \( \frac{9}{16} \).