Questions: An insulated cup contains 75.0 g of water at 24°C. A 26.00 g sample of metal at 82.25°C is added. The final temperature of water and metal is 28.34°C. What is the specific heat of the metal?

Solution Steps

• Mass of water, \( m_{\text{water}} = 75.0 \, \text{g} \)
• Initial temperature of water, \( T_{\text{initial, water}} = 24^{\circ} \mathrm{C} \)
• Final temperature of water and metal, \( T_{\text{final}} = 28.34^{\circ} \mathrm{C} \)
• Mass of metal, \( m_{\text{metal}} = 26.00 \, \text{g} \)
• Initial temperature of metal, \( T_{\text{initial, metal}} = 82.25^{\circ} \mathrm{C} \)
• Specific heat of water, \( c_{\text{water}} = 4.18 \, \text{J/g} \cdot \text{°C} \)

Use the formula for heat transfer: \[ q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial, water}}) \] \[ q_{\text{water}} = 75.0 \, \text{g} \cdot 4.18 \, \text{J/g} \cdot \text{°C} \cdot (28.34^{\circ} \mathrm{C} - 24^{\circ} \mathrm{C}) \] \[ q_{\text{water}} = 75.0 \cdot 4.18 \cdot 4.34 \] \[ q_{\text{water}} = 1360.77 \, \text{J} \]

Since the heat lost by the metal is equal to the heat gained by the water: \[ q_{\text{metal}} = -q_{\text{water}} \] \[ q_{\text{metal}} = -1360.77 \, \text{J} \]

Use the formula for heat transfer for the metal: \[ q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot (T_{\text{final}} - T_{\text{initial, metal}}) \] \[ -1360.77 \, \text{J} = 26.00 \, \text{g} \cdot c_{\text{metal}} \cdot (28.34^{\circ} \mathrm{C} - 82.25^{\circ} \mathrm{C}) \] \[ -1360.77 \, \text{J} = 26.00 \, \text{g} \cdot c_{\text{metal}} \cdot (-53.91^{\circ} \mathrm{C}) \] \[ c_{\text{metal}} = \frac{-1360.77 \, \text{J}}{26.00 \, \text{g} \cdot -53.91^{\circ} \mathrm{C}} \] \[ c_{\text{metal}} = \frac{1360.77}{1401.66} \] \[ c_{\text{metal}} \approx 0.97 \, \text{J/g} \cdot \text{°C} \]

Final Answer