Two identical systems, with heat capacity at a constant volume that varies as `C_(v) = bT^(3)` (where b is a constant) are thermally isolated. Initially, one system is at a temperature 100 K and the other is at 200K. The systems are then brought to thermal contact and the combined system is allowed reach thermal equilibrium. If `T_(0)^(4)=nxx10^(8)` then what will be the value of n, where `T_(0)` is the final temperature in kelvin?

Answer

To solve the problem, we need to find the final equilibrium temperature \( T_0 \) when two identical systems with heat capacities that vary as \( C_v = bT^3 \) are brought into thermal contact. One system starts at 100 K and the other at 200 K.

### Step-by-Step Solution:

1. **Understanding the Internal Energy Change**:
The change in internal energy \( dU \) for a system at constant volume is given by:
\[ dU = C_v dT \]
For our systems, we can express this as:
\[ dU = bT^3 dT \]

2. **Calculating Internal Energy for System 1**:
For the first system (initially at 100 K), the internal energy change from 100 K to \( T_0 \) is:
\[ U_1 = \int_{100}^{T_0} bT^3 dT \]
Evaluating this integral:
\[ U_1 = b \left[ \frac{T^4}{4} \right]_{100}^{T_0} = b \left( \frac{T_0^4}{4} - \frac{100^4}{4} \right) \]
Simplifying gives:
\[ U_1 = \frac{b}{4} (T_0^4 - 100^4) \]

3. **Calculating Internal Energy for System 2**:
For the second system (initially at 200 K), the internal energy change from \( T_0 \) to 200 K is:
\[ U_2 = \int_{T_0}^{200} bT^3 dT \]
Evaluating this integral:
\[ U_2 = b \left[ \frac{T^4}{4} \right]_{T_0}^{200} = b \left( \frac{200^4}{4} - \frac{T_0^4}{4} \right) \]
Simplifying gives:
\[ U_2 = \frac{b}{4} (200^4 - T_0^4) \]

4. **Setting Internal Energies Equal**:
At thermal equilibrium, the total internal energy before contact equals the total internal energy after contact:
\[ U_1 + U_2 = 0 \]
Therefore, we can set:
\[ \frac{b}{4} (T_0^4 - 100^4) = \frac{b}{4} (200^4 - T_0^4) \]
Canceling \( \frac{b}{4} \) from both sides gives:
\[ T_0^4 - 100^4 = 200^4 - T_0^4 \]

5. **Rearranging the Equation**:
Rearranging the equation gives:
\[ 2T_0^4 = 200^4 + 100^4 \]
Now, we calculate \( 200^4 \) and \( 100^4 \):
\[ 200^4 = (2 \times 10^2)^4 = 16 \times 10^8 \]
\[ 100^4 = (10^2)^4 = 10^8 \]
Therefore:
\[ 2T_0^4 = 16 \times 10^8 + 1 \times 10^8 = 17 \times 10^8 \]

6. **Finding \( T_0^4 \)**:
Dividing both sides by 2 gives:
\[ T_0^4 = \frac{17}{2} \times 10^8 \]

7. **Identifying \( n \)**:
From the problem, we have \( T_0^4 = n \times 10^8 \). Thus:
\[ n = \frac{17}{2} = 8.5 \]

### Final Answer:
The value of \( n \) is \( 8.5 \).