Questions: Calorimetry Thermodynamics A 5.00 kg block of metal with c=650 J /(kg*C) at 80.0°C comes in contact with a 1.25 kg glass block at 20.0°C. They come to equilibrium at 63.9°C. What is the specific heat of the glass? (Unit =J /(kg*C))

Solution Steps

We are given:

• Mass of the metal block, \( m_{\text{metal}} = 5.00 \, \text{kg} \)
• Specific heat capacity of the metal, \( c_{\text{metal}} = 650 \, \text{J/(kg} \cdot \text{C)} \)
• Initial temperature of the metal, \( T_{\text{metal, initial}} = 80.0^\circ \text{C} \)
• Mass of the glass block, \( m_{\text{glass}} = 1.25 \, \text{kg} \)
• Initial temperature of the glass, \( T_{\text{glass, initial}} = 20.0^\circ \text{C} \)
• Final equilibrium temperature, \( T_{\text{final}} = 63.9^\circ \text{C} \)

The heat lost by the metal block will be equal to the heat gained by the glass block. This can be expressed as: \[ Q_{\text{lost by metal}} = Q_{\text{gained by glass}} \]

The heat lost by the metal block is given by: \[ Q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot (T_{\text{metal, initial}} - T_{\text{final}}) \] Substituting the known values: \[ Q_{\text{metal}} = 5.00 \, \text{kg} \cdot 650 \, \text{J/(kg} \cdot \text{C)} \cdot (80.0^\circ \text{C} - 63.9^\circ \text{C}) \] \[ Q_{\text{metal}} = 5.00 \cdot 650 \cdot 16.1 \] \[ Q_{\text{metal}} = 52325 \, \text{J} \]

The heat gained by the glass block is given by: \[ Q_{\text{glass}} = m_{\text{glass}} \cdot c_{\text{glass}} \cdot (T_{\text{final}} - T_{\text{glass, initial}}) \] Since \( Q_{\text{metal}} = Q_{\text{glass}} \), we can set up the equation: \[ 52325 \, \text{J} = 1.25 \, \text{kg} \cdot c_{\text{glass}} \cdot (63.9^\circ \text{C} - 20.0^\circ \text{C}) \] \[ 52325 = 1.25 \cdot c_{\text{glass}} \cdot 43.9 \]

Rearrange the equation to solve for \( c_{\text{glass}} \): \[ c_{\text{glass}} = \frac{52325}{1.25 \cdot 43.9} \] \[ c_{\text{glass}} = \frac{52325}{54.875} \] \[ c_{\text{glass}} \approx 953.4 \, \text{J/(kg} \cdot \text{C)} \]

Final Answer