I want to keep my chickens' water trough cool in the hot summer. Is it better to put a lead kettlebell in the water or ice of the same size? The idea would be to put the kettle bell in the freezer at about 20F all night to cool it down and then put it into the water trough in the morning. Seems like a very dense lead kettle bell would absorb more heat so keep the water cooler for longer?
(Note the kettle ball is sealed so the chickens don't get lead poising!)
 A: TLDR; an ice kettle bell will absorb about 10 times as much heat as a lead one of the same size when going from 20F to 70F, even though it is less than 1/12 as dense.
Let’s pick a 20kg kettle bell.
Put it in the freezer until it is 20F. How much heat will it absorb to get to 70F?
Lead kettle bell
Specific heat of lead is 129 J/kgK (joules per gram degree C). This means it takes 129 joules of heat to raise 1 kilogram of solid lead by 1 degree C (assuming no phase changes).
It will take about 17 kcalories of heat to bring the lead kettle bell from 20F to 70F.
calculation
Ice Kettle Bell
Now let’s make a kettle bell of the same size out of ice.
The volume of the 20kg lead kettle bell is about 1750 cubic centimeters.
An ice kettle bell of the same size will only weigh 1.617 kg. So much lighter!
To figure out how much heat it will take to get the ice kettle bell form 20F to 70F, we have to break it into 3 steps – warming the ice from 20F to 32F, melting ice into water (all happens at 32F), warming the water from 32F to 70F.
Step #1 – Warm Ice
The specific heat of ice is about 2093 J/kgK.
It will take about 17 kcalories to warm the ice kettle bell from 20F to 32F.
Step #2 – Melt ice
Next we need to melt the ice. The latent heat of melting of ice is 334 J/g. It takes 334 joules of heat to turn 1 gram of ice into 1 gram of liquid water.
So, it will take ~129  kcalories of heat to melt the ice kettle bell. Phase changes can absorb huge amounts of heat!
Step #3 – Warm water
Now we have to add heat to get the water from 32F to 70F.
The specific heat of liquid water is 4182 j/kgK, so it takes about 34 kcalories to warm the water from 32F to 70F.
Total amount of heat needed to turn the 20F ice kettle bell into 70F water is about 180 kilocalories!!!
Comparison
Final results:
| Material | Heat absorbed |
| - | - |
| Lead kettle bell: | 17 kcal |
| Ice kettle bell: | 180 kcal |
