A solid copper sphere (density rho and specific heat c) of radius r at an initial temperature `200K` is suspended inside a chamber whose walls are at almost `0K.` The time required for the temperature of the sphere to drop to 100K is …….

Explanation

To solve the problem of determining the time required for the temperature of a solid copper sphere to drop from 200 K to 100 K when suspended in a chamber at nearly 0 K, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Relevant Principles**:
The cooling of the sphere occurs primarily due to radiation. According to Stefan-Boltzmann law, the power radiated by a body is proportional to the fourth power of its temperature.

2. **Write the Stefan-Boltzmann Law**:
The rate of heat loss (dq/dt) from the sphere can be expressed as:
\[ \frac{dq}{dt} = \sigma A (T^4 - T_0^4) \]
where:
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the sphere,
- \( T \) is the temperature of the sphere,
- \( T_0 \) is the temperature of the surroundings (0 K).

3. **Express the Heat Loss in Terms of Specific Heat**:
The heat loss can also be expressed in terms of the mass (m), specific heat (c), and change in temperature (dT):
\[ \frac{dq}{dt} = mc \frac{dT}{dt} \]

4. **Relate Mass to Density and Volume**:
The mass of the sphere can be expressed as:
\[ m = \rho V = \rho \left(\frac{4}{3} \pi r^3\right) \]
where \( \rho \) is the density and \( V \) is the volume of the sphere.

5. **Set the Two Expressions for Heat Loss Equal**:
Equating the two expressions for heat loss gives:
\[ mc \frac{dT}{dt} = \sigma A (T^4 - T_0^4) \]
Substituting \( A = 4 \pi r^2 \):
\[ \rho \left(\frac{4}{3} \pi r^3\right) c \frac{dT}{dt} = \sigma (4 \pi r^2) (T^4 - 0) \]

6. **Simplify the Equation**:
Cancel out \( \pi \) from both sides:
\[ \frac{4}{3} \rho r^3 c \frac{dT}{dt} = 4 \sigma r^2 T^4 \]
Simplifying further:
\[ \frac{dT}{dt} = \frac{3 \sigma T^4}{\rho r c} \]

7. **Separate Variables**:
Rearranging gives:
\[ \frac{dT}{T^4} = \frac{3 \sigma}{\rho r c} dt \]

8. **Integrate Both Sides**:
Integrate from \( T = 200 \, K \) to \( T = 100 \, K \) on the left and from \( t = 0 \) to \( t = T \) on the right:
\[ \int_{200}^{100} \frac{dT}{T^4} = \frac{3 \sigma}{\rho r c} \int_{0}^{T} dt \]

The integral on the left gives:
\[ \left[-\frac{1}{3T^3}\right]_{200}^{100} = -\frac{1}{3(100)^3} + \frac{1}{3(200)^3} \]

This simplifies to:
\[ \frac{1}{3} \left(\frac{1}{100^3} - \frac{1}{200^3}\right) \]

9. **Solve for Time**:
Set the integrated left side equal to the right side:
\[ \frac{1}{3} \left(\frac{1}{100^3} - \frac{1}{200^3}\right) = \frac{3 \sigma T}{\rho r c} \]

Rearranging gives:
\[ T = \frac{1}{\frac{3 \sigma}{\rho r c} \left(\frac{1}{100^3} - \frac{1}{200^3}\right)} \]

### Final Result
The time required for the temperature of the sphere to drop from 200 K to 100 K can be expressed as:
\[ T = \frac{1}{\frac{3 \sigma}{\rho r c} \left(\frac{1}{100^3} - \frac{1}{200^3}\right)} \]