Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at `100^(@)C`. While the other one is at `0^(@)C`. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is

Options

A. less than `50^(@)C` but greater than `0^(@)C`
B. `0^(@)C`
C. `50^(@)C`
D. more than `50^(@)C`

Explanation

To solve the problem of finding the final common temperature when two identical bodies made of a material with increasing heat capacity are brought into contact, we can follow these steps:

### Step 1: Define Variables
Let:
- \( \theta \) = final common temperature
- \( S_c \) = average heat capacity of the cold body (at \( 0^\circ C \))
- \( S_h \) = average heat capacity of the hot body (at \( 100^\circ C \))

### Step 2: Set Up the Heat Transfer Equation
According to the principle of conservation of energy, the heat gained by the cold body is equal to the heat lost by the hot body. We can express this as:
\[ S_c (\theta - 0) = S_h (100 - \theta) \]

### Step 3: Rearrange the Equation
From the equation, we can rearrange it to isolate \( \theta \):
\[ S_c \theta = S_h (100 - \theta) \]
\[ S_c \theta + S_h \theta = 100 S_h \]
\[ \theta (S_c + S_h) = 100 S_h \]

### Step 4: Solve for \( \theta \)
Now, we can solve for \( \theta \):
\[ \theta = \frac{100 S_h}{S_c + S_h} \]

### Step 5: Analyze the Ratio of Heat Capacities
Since the heat capacity increases with temperature, we know that \( S_c < S_h \). Therefore, the ratio \( \frac{S_c}{S_h} < 1 \).

### Step 6: Substitute and Simplify
We can express \( \theta \) in terms of the ratio:
\[ \theta = \frac{100}{1 + \frac{S_c}{S_h}} \]
Since \( \frac{S_c}{S_h} < 1 \), we can conclude that:
\[ 1 + \frac{S_c}{S_h} < 2 \]
Thus:
\[ \theta > \frac{100}{2} = 50^\circ C \]

### Conclusion
The final common temperature \( \theta \) is greater than \( 50^\circ C \).