A small object is placed at the center of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time 𝑡 = 0, the temperature of the object is 200 K. The temperature of the object becomes 100 K at t=𝑡1 and 50 K at 𝑡 = 𝑡2. Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio (𝑡2/𝑡1) is ___.

 To determine the ratio 

$\frac{t_2}{t_1}$, we need to analyze the cooling process of the object in the evacuated hollow spherical container. Since both the object and the container are ideal black bodies, the cooling occurs solely through radiation.

1. Understanding the Problem

• Object: Initially at 200 K, cools to 100 K at time $t_1$ and to 50 K at time $t_2$.

• Container: Maintained at 0 K.

• Assumptions:

  • Both the object and the container are ideal black bodies.

  • The heat capacity of the object does not depend on temperature.

  • The container is evacuated, so heat transfer occurs only by radiation.

2. Stefan-Boltzmann Law

The rate at which an object radiates energy (power) is given by the Stefan-Boltzmann law:

$$P=\sigma A{T}^4$$

where:

• $P$ is the power radiated,

• $\sigma$ is the Stefan-Boltzmann constant,

• $A$ is the surface area of the object,

• $T$ is the absolute temperature of the object.

Since the container is at 0 K, it does not radiate back to the object, so the net power radiated by the object is:

$$P_{\text{net}}=\sigma A{T}^4$$

3. Cooling Rate and Heat Capacity

The heat capacity $C$ of the object relates the change in temperature to the heat lost:

$$C\frac{dT}{dt}=-P_{\text{net}}=-\sigma A{T}^4$$

Rearranging:

$$\frac{dT}{dt}=-\frac{\sigma A}{C}{T}^4$$

Let $k=\frac{\sigma A}{C}$, so:

$$\frac{dT}{dt}=-k{T}^4$$

4. Solving the Differential Equation

Separate variables and integrate:

$$\int \frac{1}{T^4}dT=-k\int dt$$$$\int T^{-4}dT=-k\int dt$$$$\frac{T^{-3}}{-3}=-kt+\text{constant}$$$$\frac{1}{3T^3}=kt+\text{constant}$$

5. Applying Initial Conditions

At $t=0$, $T=200$ K:

$$\frac{1}{3(200{)}^3}=k\cdot 0+\text{constant}⟹\text{constant}=\frac{1}{3(200{)}^3}$$

So, the equation becomes:

$$\frac{1}{3T^3}=kt+\frac{1}{3(200{)}^3}$$

6. Finding $t_1$ and $t_2$

We need to find the times $t_1$ and $t_2$ when the temperature is 100 K and 50 K, respectively.

For $t_1$ (when $T=100$ K):

$$\frac{1}{3(100{)}^3}=k{t}_1+\frac{1}{3(200{)}^3}$$

Solving for $t_1$:

$$k{t}_1=\frac{1}{3(100{)}^3}-\frac{1}{3(200{)}^3}$$$$k{t}_1=\frac{1}{3\times 10^6}-\frac{1}{3\times 8\times 10^6}=\frac{1}{3\times 10^6}(1-\frac{1}{8})=\frac{1}{3\times 10^6}\times \frac{7}{8}=\frac{7}{24\times 10^6}$$$$t_1=\frac{7}{24\times 10^{6}k}$$

For $t_2$ (when $T=50$ K):

$$\frac{1}{3(50{)}^3}=k{t}_2+\frac{1}{3(200{)}^3}$$

Solving for $t_2$:

$$k{t}_2=\frac{1}{3(50{)}^3}-\frac{1}{3(200{)}^3}$$$$k{t}_2=\frac{1}{3\times 125\times 10^3}-\frac{1}{3\times 8\times 10^6}=\frac{1}{375\times 10^3}-\frac{1}{24\times 10^6}$$$$k{t}_2=\frac{1}{375\times 10^3}(1-\frac{1}{64})=\frac{1}{375\times 10^3}\times \frac{63}{64}=\frac{63}{24\times 10^6}$$$$t_2=\frac{63}{24\times 10^{6}k}$$

7. Calculating the Ratio $\frac{t_2}{t_1}$

Now, we can find the ratio:

$$\frac{t_2}{t_1}=\frac{\frac{63}{24\times 10^{6}k}}{\frac{7}{24\times 10^{6}k}}=\frac{63}{7}=9$$

8. Conclusion

The ratio $\frac{t_2}{t_1}$ is 9.

Final Answer: $\frac{t_2}{t_1}=9$