Temperature of 100 g of water in a thermoflask remains fixed for a pretty long time at `50^(@)C`. An equal mass of sand at `20^(@)C` is poured in the flask and shaken for some time so that the temperature of the mixture is `40^(@)`. Now the experiment is repeated with 100g of a liquid at `50^(@) C` and an equal amount of sand at `20^(@)C` when the temperature of the mixture is found to be `30^(@)C` The specific heat of the liquid `(in KJ//KgxK)` is

Options

A. `1.05`
B. `2.10`
C. `4.20`
D. `1.951`

Explanation

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the hotter substance will be equal to the heat gained by the cooler substance.

### Step-by-Step Solution:

1. **Identify the known values:**
- Mass of water (m_w) = 100 g = 0.1 kg
- Initial temperature of water (T_w_initial) = 50°C
- Final temperature of mixture (T_final) = 40°C
- Initial temperature of sand (T_s_initial) = 20°C
- Mass of sand (m_s) = 100 g = 0.1 kg
- Specific heat of water (c_w) = 4.2 kJ/kg·K

2. **Calculate the heat lost by water:**
\[ Q_w = m_w \cdot c_w \cdot (T_w_initial - T_final) \]
\[ Q_w = 0.1 \cdot 4.2 \cdot (50 - 40) = 0.1 \cdot 4.2 \cdot 10 = 4.2 \text{ kJ} \]

3. **Calculate the heat gained by sand:**
\[ Q_s = m_s \cdot c_s \cdot (T_final - T_s_initial) \]
\[ Q_s = 0.1 \cdot c_s \cdot (40 - 20) = 0.1 \cdot c_s \cdot 20 = 2c_s \text{ kJ} \]

4. **Set the heat lost by water equal to the heat gained by sand:**
\[ Q_w = Q_s \]
\[ 4.2 = 2c_s \]
\[ c_s = \frac{4.2}{2} = 2.1 \text{ kJ/kg·K} \]

5. **Now repeat the experiment with liquid:**
- Mass of liquid (m_l) = 100 g = 0.1 kg
- Initial temperature of liquid (T_l_initial) = 50°C
- Final temperature of mixture (T_final) = 30°C

6. **Calculate the heat lost by liquid:**
\[ Q_l = m_l \cdot c_l \cdot (T_l_initial - T_final) \]
\[ Q_l = 0.1 \cdot c_l \cdot (50 - 30) = 0.1 \cdot c_l \cdot 20 = 2c_l \text{ kJ} \]

7. **Set the heat lost by liquid equal to the heat gained by sand:**
\[ Q_l = Q_s \]
\[ 2c_l = 2c_s \]
\[ c_l = c_s \]

8. **Substituting the value of \(c_s\):**
\[ c_l = 2.1 \text{ kJ/kg·K} \]

9. **Final Calculation:**
The specific heat of the liquid is therefore \(c_l = 2.1 \text{ kJ/kg·K}\).

### Conclusion:
The specific heat of the liquid is **2.1 kJ/kg·K**.