(1pts) The heat pack should be able to raise the temperature from \( 15^{\circ} \mathrm{C} \) to between \( 36.0^{\circ} \mathrm{C} \) and \( 40.0^{\circ} \mathrm{C} \). If dissolving 5.0 g of \( \mathrm{CaCl}_{2} \) in 40.0 mL of water produces a temperature increase of \( 22-25^{\circ} \mathrm{C} \), estimate the volume of water required to achieve a temperature change within this range.

To estimate the volume of water required to achieve a temperature change within the range of \( 22-25^{\circ} \mathrm{C} \) when dissolving \( 5.0 \, \text{g} \) of \( \mathrm{CaCl}_{2} \), we can use the concept of heat transfer and the specific heat capacity of water. 1. **Calculate the heat released by dissolving \( 5.0 \, \text{g} \) of \( \mathrm{CaCl}_{2} \)**: The heat released (\( q \)) can be calculated using the formula: \[ q = m \cdot C \cdot \Delta T \] where: - \( m \) is the mass of water (in grams), - \( C \) is the specific heat capacity of water (\( 4.18 \, \text{J/g} \cdot \text{°C} \)), - \( \Delta T \) is the temperature change. 2. **Determine the temperature change**: The temperature change (\( \Delta T \)) when dissolving \( 5.0 \, \text{g} \) of \( \mathrm{CaCl}_{2} \) in \( 40.0 \, \text{mL} \) of water is given as \( 22-25^{\circ} \mathrm{C} \). We can take the average temperature change for our calculations: \[ \Delta T = \frac{22 + 25}{2} = 23.5^{\circ} \mathrm{C} \] 3. **Calculate the heat released**: The heat released when dissolving \( 5.0 \, \text{g} \) of \( \mathrm{CaCl}_{2} \) in \( 40.0 \, \text{mL} \) of water can be calculated as follows: \[ q = m \cdot C \cdot \Delta T \] The mass of water in \( 40.0 \, \text{mL} \) is approximately \( 40.0 \, \text{g} \) (since the density of water is \( 1 \, \text{g/mL} \)): \[ q = 40.0 \, \text{g} \cdot 4.18 \, \text{J/g} \cdot \text{°C} \cdot 23.5^{\circ} \mathrm{C} \] \[ q \approx 40.0 \cdot 4.18 \cdot 23.5 \approx 3925.4 \, \text{J} \] 4. **Determine the volume of water required for the desired temperature change**: We want to find the volume of water (\( V \)) that will achieve a temperature change of \( 22-25^{\circ} \mathrm{C} \) using the same amount of heat released by \( 5.0 \, \text{g} \) of \( \mathrm{CaCl}_{2} \): \[ q = V \cdot C \cdot \Delta T \] Rearranging for \( V \): \[ V = \frac{q}{C \cdot \Delta T} \] We can use the average temperature change of \( 23.5^{\circ} \mathrm{C} \): \[ V = \frac{3925.4 \, \text{J}}{4.18 \, \text{J/g} \cdot \text{°C} \cdot 23.5^{\circ} \mathrm{C}} \] \[ V \approx \frac{3925.4}{4.18 \cdot 23.5} \approx \frac{3925.4}{98.13} \approx 40.0 \, \text{g} \] 5. **Convert grams to milliliters**: Since the density of water is \( 1 \, \text{g/mL} \), the volume of water required is approximately: \[ V \approx 40.0 \, \text{mL} \] Thus, the volume of water required to achieve a temperature change within the range of \( 22-25^{\circ} \mathrm{C} \) when dissolving \( 5.0 \, \text{g} \) of \( \mathrm{CaCl}_{2}