A piece of glass has a temperature of `83.0^(@)C`. Liquid that has a temperature of `43.0^(@)C` is poured over the glass, completely covering it, and the temperature at equilibrium is `53.0^(@)C` . The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible , determine the specific heat capacity of the liquid.

Options

A. `280 J //( kg. ""^(@)C )`
B. `1300 J // ( kg. ""^(@)C )`
C. `2500J//(kg. ""^(@)C )`
D. `3400 J //kg ( kg. ""^(@)C )`

Answer

To determine the specific heat capacity of the liquid, we can use the principle of conservation of energy, which states that the heat lost by the glass will be equal to the heat gained by the liquid.

### Step-by-Step Solution:

1. **Identify the known values:**
- Initial temperature of glass, \( T_{g_i} = 83.0^\circ C \)
- Initial temperature of liquid, \( T_{l_i} = 43.0^\circ C \)
- Final equilibrium temperature, \( T_f = 53.0^\circ C \)
- Specific heat capacity of glass, \( c_g = 840 \, \text{J/kg}^\circ C \)
- Let the mass of glass and liquid be \( m \) (since they are the same).

2. **Calculate the heat lost by the glass:**
\[ Q_g = m \cdot c_g \cdot (T_{g_i} - T_f) \]
Substituting the known values:
\[ Q_g = m \cdot 840 \, \text{J/kg}^\circ C \cdot (83.0 - 53.0) \]
\[ Q_g = m \cdot 840 \, \text{J/kg}^\circ C \cdot 30.0 \]
\[ Q_g = m \cdot 25200 \, \text{J} \]

3. **Calculate the heat gained by the liquid:**
\[ Q_l = m \cdot c_l \cdot (T_f - T_{l_i}) \]
Here, \( c_l \) is the specific heat capacity of the liquid which we need to find. Substituting the known values:
\[ Q_l = m \cdot c_l \cdot (53.0 - 43.0) \]
\[ Q_l = m \cdot c_l \cdot 10.0 \]
\[ Q_l = m \cdot 10.0 \cdot c_l \, \text{J} \]

4. **Set the heat lost by the glass equal to the heat gained by the liquid:**
\[ Q_g = Q_l \]
\[ m \cdot 25200 = m \cdot 10.0 \cdot c_l \]

5. **Cancel the mass \( m \) from both sides (since \( m \neq 0 \)):**
\[ 25200 = 10.0 \cdot c_l \]

6. **Solve for \( c_l \):**
\[ c_l = \frac{25200}{10.0} \]
\[ c_l = 2520 \, \text{J/kg}^\circ C \]

### Final Answer:
The specific heat capacity of the liquid is \( 2520 \, \text{J/kg}^\circ C \).