How many Btu must be removed from one pound of water at 200F for it to end up as ice at 30F?

To calculate the total BTUs removed to convert one pound of water at 200°F to ice at 30°F, you need to consider two processes: cooling the water from 200°F to 32°F and then freezing it to ice at 32°F, followed by cooling the ice to 30°F.

1. Cooling water from 200°F to 32°F: This requires removing about 168 BTUs (1 BTU per pound per °F * (200°F - 32°F)).
2. Freezing the water at 32°F: This requires removing 80 BTUs (the latent heat of fusion).
3. Cooling the ice from 32°F to 30°F: This requires removing 2 BTUs (1 BTU per pound per °F * (32°F - 30°F)).

Adding these together: 168 BTUs + 80 BTUs + 2 BTUs = 250 BTUs must be removed.