Answer:
The amount of heat that must be removed from steam at 100 degrees Celsius to get ice at -273 degrees Celsius can be calculated using the formula for the heat of fusion of water. The heat of fusion of water is approximately 333.55 joules per gram, which means that it takes 333.55 joules of energy to convert 1 gram of liquid water at 0 degrees Celsius to 1 gram of ice at 0 degrees Celsius.
To convert 0.000010 kg of steam at 100 degrees Celsius to ice at -273 degrees Celsius, we must first convert the steam to liquid water at 100 degrees Celsius, which requires the addition of heat. The heat required to do this can be calculated using the specific heat capacity of water, which is approximately 4.186 joules per gram per degree Celsius. The specific heat capacity tells us how much heat is required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
To convert 0.000010 kg of steam at 100 degrees Celsius to liquid water at 100 degrees Celsius, we would need to add (0.000010 kg) * (100 degrees Celsius) * (4.186 joules/gram/degree Celsius) = 0.004186 joules of heat.
Once the steam has been converted to liquid water, we can then convert the water to ice at -273 degrees Celsius. To do this, we would need to remove (0.000010 kg) * (333.55 joules/gram) = 0.033355 joules of heat.
In total, we would need to remove 0.004186 joules + 0.033355 joules = 0.037541 joules of heat from 0.000010 kg of steam at 100 degrees Celsius to get ice at -273 degrees Celsius.
Step-by-step explanation: