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 freeze the water, you need to consider two main stages. First, you'll need to lower the temperature of the water from \( 50^{\circ} \mathrm{F} \) to \( 32^{\circ} \mathrm{F} \), and then you'll need to change the state of that water from liquid to solid. The specific heat of water is approximately 1 BTU/lb/°F, so to cool it down to freezing involves removing: \[ 5 \, \text{lbs} \times (50 - 32) \, \text{°F} = 5 \times 18 = 90 \, \text{BTUs} .\] Once it reaches \( 32^{\circ} \mathrm{F} \), you need to remove the latent heat of fusion, which is about 144 BTUs per pound of ice formed. Thus for 5 lbs of water: \[ 5 \, \text{lbs} \times 144 \, \text{BTUs/lb} = 720 \, \text{BTUs} .\] Adding these together gives you the total: \[ 90 \, \text{BTUs} + 720 \, \text{BTUs} = 810 \, \text{BTUs} .\] So, the final answer is (A) 810.