Heat Removal from Ice: How Much Energy?

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The discussion centers on the energy calculations involved in transitioning water to ice and back. It confirms that removing 1200 BTUs is necessary to convert one gallon of water at 32.1 degrees F into ice. When the ice is placed in a 0 degrees F environment, it will release an additional 266 BTUs to reach thermal equilibrium. To melt the ice back into water at 32.1 degrees F, 266 BTUs must first be added to raise the temperature to the transition point, followed by an additional 1200 BTUs for melting. The conversation highlights the lower specific heat capacity of ice compared to water, which affects the energy calculations.
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I want to start with several assumptions:

1. I have one gallon of water at 32.1 degrees F
2. It is correct that in order to lower the temperature of one gallon of water 1 degree F, approximately 8.34 BTU's of heat energy must be removed from that water.
3. Latent heat of fusion requires that approximately 1200 BTU's be removed from the water to convert it into ice.

Let's assume that I remove that 1200 BTU's of heat from the water and it then transitions into ice. Now let's further assume I place that gallon of ice, which is at 32 degrees F, into a room that is maintained at 0 degrees F by refrigeration.

Am I correct in assuming the following:

1. The ice will release an additional 266 BTU's of energy to reach equilibrium temperature with the room, i.e., 0 degrees F? Of course, this assumes the refrigeration equipment is capable of removing that heat from the room and maintaining 0 degrees F.

2. In order to melt that ice back into 32.1 degree F water, I will need to add 266 BTU's to that gallon of ice to bring it to its transition temperature of 32 degrees. I will then need to add 1200 BTU's of heat energy to cause that ice to melt back into water that is just above the freezing point?

Thanks for any help.
 
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Not quite, the specific heat capacity of ice (and steam oddly enough) is less than half that of water. So it only takes 2050J/Kg to raise ice 1degC rather than 4180J/kg for water (same ratio in imperial units).

Otherwise you're correct.
 
Thanks for the prompt and helpful reply. Good to know I had not forgotten everything from my HS and college physics 20-some years ago.

Thanks again.
 
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