The mass, specific heat capacity and the temperature of a solid are 1000g,(1/2) ((cal)/g) and 80 degrees C respectively. The mass of the liquid and the calorimeter are 900g` and 200g. Initially, both are at room temperature 20 degrees C. Both calorimeter and the solid are made of the same material. In the steady state, temperature of mixture is 40 degrees C, then specific heat capacity of the unknown liquid is;

Explanation

To solve the problem, we need to apply the principle of conservation of energy, which states that the total heat lost by the solid and the calorimeter must equal the total heat gained by the liquid.

### Step-by-step Solution:

1. **Identify Given Values**:
- Mass of the solid, \( m_s = 1000 \, \text{g} \)
- Specific heat capacity of the solid, \( c_s = \frac{1}{2} \, \text{cal/g°C} \)
- Initial temperature of the solid, \( T_{s_i} = 80 \, \text{°C} \)
- Mass of the liquid, \( m_l = 900 \, \text{g} \)
- Mass of the calorimeter, \( m_c = 200 \, \text{g} \)
- Initial temperature of the liquid and calorimeter, \( T_{l_i} = T_{c_i} = 20 \, \text{°C} \)
- Final temperature of the mixture, \( T_f = 40 \, \text{°C} \)

2. **Calculate Heat Lost by the Solid**:
The heat lost by the solid can be calculated using the formula:
\[ Q_s = m_s \cdot c_s \cdot (T_{s_i} - T_f) \]
Substituting the values:
\[ Q_s = 1000 \, \text{g} \cdot \frac{1}{2} \, \text{cal/g°C} \cdot (80 - 40) \]
\[ Q_s = 1000 \cdot \frac{1}{2} \cdot 40 = 20000 \, \text{cal} \]

3. **Calculate Heat Gained by the Calorimeter**:
The heat gained by the calorimeter is:
\[ Q_c = m_c \cdot c_s \cdot (T_f - T_{c_i}) \]
Substituting the values:
\[ Q_c = 200 \, \text{g} \cdot \frac{1}{2} \, \text{cal/g°C} \cdot (40 - 20) \]
\[ Q_c = 200 \cdot \frac{1}{2} \cdot 20 = 2000 \, \text{cal} \]

4. **Set Up the Heat Gain Equation for the Liquid**:
The heat gained by the liquid can be expressed as:
\[ Q_l = m_l \cdot c_l \cdot (T_f - T_{l_i}) \]
Substituting the values:
\[ Q_l = 900 \, \text{g} \cdot c_l \cdot (40 - 20) \]
\[ Q_l = 900 \cdot c_l \cdot 20 \]

5. **Apply Conservation of Energy**:
According to the conservation of energy:
\[ Q_s + Q_c + Q_l = 0 \]
Substituting the values:
\[ -20000 + 2000 + 900 \cdot c_l \cdot 20 = 0 \]
Simplifying this gives:
\[ 900 \cdot c_l \cdot 20 = 20000 - 2000 \]
\[ 900 \cdot c_l \cdot 20 = 18000 \]

6. **Solve for Specific Heat Capacity of the Liquid**:
\[ c_l = \frac{18000}{900 \cdot 20} \]
\[ c_l = \frac{18000}{18000} = 1 \, \text{cal/g°C} \]

### Final Answer:
The specific heat capacity of the unknown liquid is \( 1 \, \text{cal/g°C} \).