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 need to 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 1: Write down the given data
- Mass of metal X, \( m_X = 5 \, \text{g} \)
- Specific heat of metal X, \( s_X = 0.2 \, \text{cal/g}^\circ C \)
- Initial temperature of metal X, \( T_{iX} = 40^\circ C \)
- Final equilibrium temperature in calorimeter A, \( T_A = 22^\circ C \)
- Initial temperature of water, \( T_{w} = 20^\circ C \)

- Mass of metal Y, \( m_Y = 5 \, \text{g} \)
- Initial temperature of metal Y, \( T_{iY} = 40^\circ C \)
- Final equilibrium temperature in calorimeter B, \( T_B = 23^\circ C \)

### Step 2: Calculate the temperature changes
- Temperature change for metal X:
\[ \Delta T_X = T_{iX} - T_A = 40^\circ C - 22^\circ C = 18^\circ C \]

- Temperature change for metal Y:
\[ \Delta T_Y = T_{iY} - T_B = 40^\circ C - 23^\circ C = 17^\circ C \]

- Temperature change for water in calorimeter A:
\[ \Delta T_{wA} = T_A - T_{w} = 22^\circ C - 20^\circ C = 2^\circ C \]

- Temperature change for water in calorimeter B:
\[ \Delta T_{wB} = T_B - T_{w} = 23^\circ C - 20^\circ C = 3^\circ C \]

### Step 3: Set up the heat transfer equations
For calorimeter A (with metal X):
\[ \text{Heat lost by metal X} = \text{Heat gained by water} \]
\[ m_X \cdot s_X \cdot \Delta T_X = m_W \cdot s_W \cdot \Delta T_{wA} \]
Where \( m_W \) is the mass of water and \( s_W = 1 \, \text{cal/g}^\circ C \) (specific heat of water).

Substituting the known values:
\[ 5 \cdot 0.2 \cdot 18 = m_W \cdot 1 \cdot 2 \]
\[ 18 = 2m_W \implies m_W = 9 \, \text{g} \]

For calorimeter B (with metal Y):
\[ \text{Heat lost by metal Y} = \text{Heat gained by water} \]
\[ m_Y \cdot s_Y \cdot \Delta T_Y = m_W \cdot s_W \cdot \Delta T_{wB} \]
Substituting the known values:
\[ 5 \cdot s_Y \cdot 17 = 9 \cdot 1 \cdot 3 \]
\[ 85 = 5s_Y \implies s_Y = \frac{85}{5} = 17 \, \text{cal/g}^\circ C \]

### Final Answer
The specific heat of metal Y is \( 17 \, \text{cal/g}^\circ C \).

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