7. 15 grams of Li is placed into a 500 gram copper calorimeter containing 150 grams of water. The calorimeter experiences a temperature change of 20 degrees Celsius. \( \begin{array}{ll}15 & \text { a) Determine the energy produced by this sample. }\end{array} \)

To determine the energy produced, we can use the specific heat formula: \[ Q = mc\Delta T \] Where \( Q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. The specific heat capacity of water is approximately \( 4.18 \, \text{J/g°C} \). First, we find the energy produced by the water: \[ m = 150 \, \text{g}, \, \Delta T = 20 \, \text{°C} \] Thus, \[ Q_{\text{water}} = 150 \, \text{g} \times 4.18 \, \text{J/g°C} \times 20 \, \text{°C} \] Now calculating this gives: \[ Q_{\text{water}} = 150 \times 4.18 \times 20 = 12540 \, \text{J} \] Now, let’s calculate the energy exchanged with the calorimeter. The specific heat capacity of copper is approximately \( 0.385 \, \text{J/g°C} \). For the calorimeter: \[ m = 500 \, \text{g} \] Thus, \[ Q_{\text{copper}} = 500 \, \text{g} \times 0.385 \, \text{J/g°C} \times 20 \, \text{°C} \] \[ Q_{\text{copper}} = 500 \times 0.385 \times 20 = 3850 \, \text{J} \] Total energy produced: \[ Q_{\text{total}} = Q_{\text{water}} + Q_{\text{copper}} \] \[ Q_{\text{total}} = 12540 \, \text{J} + 3850 \, \text{J} = 16400 \, \text{J} \] So, the energy produced by the sample is \( 16400 \, \text{J} \).