Two identical calorimeters A and B contain an equal quantity of water at `20^(@)C`. A 5 g piece of metal X of specific heat `0.2" cal g"^(-1).^(@)C^(-1)` is dropped into A and 5 g piece of metal Y is dropped into B. The equilibrium temperature in A is `22^(@)C` and that in B is `23^(@)C`. The intial temperature of both the metals was `40^(@)C`. The specific heat of metal Y `("in cal g"^(-1).^(@)C^(-1))` is

Options

A. `(27)/(85)`
B. `(54)/(85)`
C. `(81)/(85)`
D. 0.4

Answer

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the metal will be equal to the heat gained by the water in the calorimeter.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Mass of metal X (m_X) = 5 g
- Specific heat of metal X (S_X) = 0.2 cal/g°C
- Initial temperature of metals (T_initial) = 40°C
- Final temperature in calorimeter A (T_A) = 22°C
- Initial temperature of water in calorimeter A (T_water) = 20°C
- Mass of water (m_water) = m (unknown but equal for both calorimeters)
- Specific heat of water (S_water) = 1 cal/g°C
- Final temperature in calorimeter B (T_B) = 23°C
- Mass of metal Y (m_Y) = 5 g
- Specific heat of metal Y (S_Y) = ? (to be determined)

2. **Calculate Heat Lost by Metal X in Calorimeter A:**
\[ Q_X = m_X \cdot S_X \cdot (T_initial - T_A) \]
\[ Q_X = 5 \cdot 0.2 \cdot (40 - 22) = 5 \cdot 0.2 \cdot 18 = 18 \text{ cal} \]

3. **Calculate Heat Gained by Water in Calorimeter A:**
\[ Q_{waterA} = m \cdot S_{water} \cdot (T_A - T_water) \]
\[ Q_{waterA} = m \cdot 1 \cdot (22 - 20) = m \cdot 2 \text{ cal} \]

4. **Set Heat Lost by Metal X Equal to Heat Gained by Water:**
\[ Q_X = Q_{waterA} \]
\[ 18 = m \cdot 2 \implies m = 9 \text{ g} \]

5. **Calculate Heat Lost by Metal Y in Calorimeter B:**
\[ Q_Y = m_Y \cdot S_Y \cdot (T_initial - T_B) \]
\[ Q_Y = 5 \cdot S_Y \cdot (40 - 23) = 5 \cdot S_Y \cdot 17 \]

6. **Calculate Heat Gained by Water in Calorimeter B:**
\[ Q_{waterB} = m \cdot S_{water} \cdot (T_B - T_water) \]
\[ Q_{waterB} = 9 \cdot 1 \cdot (23 - 20) = 9 \cdot 3 = 27 \text{ cal} \]

7. **Set Heat Lost by Metal Y Equal to Heat Gained by Water:**
\[ Q_Y = Q_{waterB} \]
\[ 5 \cdot S_Y \cdot 17 = 27 \]

8. **Solve for Specific Heat of Metal Y (S_Y):**
\[ S_Y = \frac{27}{5 \cdot 17} = \frac{27}{85} \text{ cal/g°C} \]

### Final Answer:
The specific heat of metal Y is \(\frac{27}{85}\) cal/g°C.