Answer :
The specific heat of the block is 0.867 J/g ° C.
To calculate the specific heat of the block, we use the principle of conservation of energy, where the heat lost by the block equals the heat gained by the water and the calorimeter.
Given Data:
- Mass of the block, [tex]\( m_{\text{block}} = 21.92 \, \text{g} \)[/tex]
- Mass of water and calorimeter, [tex]\( m_{\text{water+cal}} = 53.07 \, \text{g} \)[/tex]
- Mass of the empty calorimeter, [tex]\( m_{\text{cal}} = 4.34 \, \text{g} \)[/tex]
- Mass of water, [tex]\( m_{\text{water}} = 53.07 \, \text{g} - 4.34 \, \text{g} = 48.73 \, \text{g} \)[/tex]
- Initial temperature of the block (boiling water), [tex]\( T_{\text{initial, block}} = 97.6 \, ^\circ\text{C} \)[/tex]
- Initial temperature of the water and calorimeter, [tex]\( T_{\text{initial, water}} = 19.8 \, ^\circ\text{C} \)[/tex]
- Final temperature of the system,[tex]\( T_{\text{final}} = 25.9 \, ^\circ\text{C} \)[/tex]
- Specific heat of water, [tex]\( c_{\text{water}} = 4.184 \, \text{J/g}^\circ\text{C} \)[/tex]
Heat Gained by Water and Calorimeter:
1. Water:
[tex]\[ q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial, water}}) \]\[ q_{\text{water}} = 48.73 \, \text{g} \cdot 4.184 \, \text{J/g}^\circ\text{C} \cdot (25.9 \, ^\circ\text{C} - 19.8 \, ^\circ\text{C}) \]\[ q_{\text{water}} = 48.73 \, \text{g} \cdot 4.184 \, \text{J/g}^\circ\text{C} \cdot 6.1 \, ^\circ\text{C} \]\[ q_{\text{water}} = 1241.88 \, \text{J} \][/tex]
2. Calorimeter:
Assuming the specific heat of the calorimeter is similar to water:
[tex]\[ q_{\text{cal}} = m_{\text{cal}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial, water}}) \]\[ q_{\text{cal}} = 4.34 \, \text{g} \cdot 4.184 \, \text{J/g}^\circ\text{C} \cdot (25.9 \, ^\circ\text{C} - 19.8 \, ^\circ\text{C}) \]\[ q_{\text{cal}} = 4.34 \, \text{g} \cdot 4.184 \, \text{J/g}^\circ\text{C} \cdot 6.1 \, ^\circ\text{C} \]\[ q_{\text{cal}} = 110.63 \, \text{J} \][/tex]
Total Heat Gained:
[tex]\[ q_{\text{total, gained}} = q_{\text{water}} + q_{\text{cal}} \]\[ q_{\text{total, gained}} = 1241.88 \, \text{J} + 110.63 \, \text{J} \]\[ q_{\text{total, gained}} = 1352.51 \, \text{J} \][/tex]
Heat Lost by the Block:
[tex]\[ q_{\text{block}} = -q_{\text{total, gained}} \]\[ q_{\text{block}} = -1352.51 \, \text{J} \][/tex]
Specific Heat of the Block:
[tex]\[ q_{\text{block}} = m_{\text{block}} \cdot c_{\text{block}} \cdot (T_{\text{final}} - T_{\text{initial, block}}) \]\[ -1352.51 \, \text{J} = 21.92 \, \text{g} \cdot c_{\text{block}} \cdot (25.9 \, ^\circ\text{C} - 97.6 \, ^\circ\text{C}) \][/tex]
[tex]\[ -1352.51 \, \text{J} = 21.92 \, \text{g} \cdot c_{\text{block}} \cdot (-71.7 \, ^\circ\text{C}) \]\[ c_{\text{block}} = \frac{-1352.51 \, \text{J}}{21.92 \, \text{g} \cdot (-71.7 \, ^\circ\text{C})} \][/tex]
[tex]\[ c_{\text{block}} = \frac{1352.51 \, \text{J}}{21.92 \, \text{g} \cdot 71.7 \, ^\circ\text{C}} \]\[ c_{\text{block}} \approx 0.867 \, \text{J/g}^\circ\text{C} \][/tex]
Therefore, the specific heat of the block is approximately [tex]\( 0.867 \, \text{J/g}^\circ\text{C} \).[/tex]