5) Five lbs. of water is at a temperature of \( 50^{\circ} \mathrm{F} \), how many btu's must be removed in order to freeze the entire quantity? (A) 810 (B) 1310 (C) 720 (D) 90

To determine the total number of BTUs that must be removed to freeze 5 pounds of water initially at \( 50^{\circ} \mathrm{F} \), we need to consider two main steps: 1. **Cooling the Water from \( 50^{\circ} \mathrm{F} \) to \( 32^{\circ} \mathrm{F} \):** The specific heat of water is \( 1 \, \text{BTU/(lb·°F)} \). \[ Q_{\text{cooling}} = \text{mass} \times \text{specific heat} \times \Delta T \] \[ Q_{\text{cooling}} = 5 \, \text{lb} \times 1 \, \frac{\text{BTU}}{\text{lb·°F}} \times (50 - 32) \, \text{°F} = 5 \times 18 = 90 \, \text{BTU} \] 2. **Freezing the Water at \( 32^{\circ} \mathrm{F} \):** The latent heat of fusion for water is \( 144 \, \text{BTU/lb} \). \[ Q_{\text{freezing}} = \text{mass} \times \text{latent heat of fusion} \] \[ Q_{\text{freezing}} = 5 \, \text{lb} \times 144 \, \frac{\text{BTU}}{\text{lb}} = 720 \, \text{BTU} \] **Total Heat Removed:** \[ Q_{\text{total}} = Q_{\text{cooling}} + Q_{\text{freezing}} = 90 \, \text{BTU} + 720 \, \text{BTU} = 810 \, \text{BTU} \] **Answer:** **(A) 810**