How many calories would be needed to convert 8.0g of ice at 0C to steam at 100C?
To calculate the amount of calories needed to convert the given quantity of ice at 0°C to steam at 100°C, we need to consider the processes of heating the ice to its melting point, melting it, heating the resulting water to its boiling point, and vaporizing it.
Here’s how you can calculate it:
1. Calculate the energy required to heat the ice from 0°C to its melting point at 0°C:
- The specific heat capacity of ice is 2.09 calories (g°C)^-1.
- So, the energy required to heat 8.0g of ice by 1°C is (2.09 calories/g°C) * 8.0g * 1°C = 16.72 calories.
2. Calculate the energy required to melt the ice at its melting point:
- The heat of fusion (or latent heat) of ice is 79.7 calories/g.
- So, the energy required to melt 8.0g of ice is (79.7 calories/g) * 8.0g = 637.6 calories.
3. Calculate the energy required to heat the resulting water from 0°C to its boiling point at 100°C:
- The specific heat capacity of water is 4.18 calories (g°C)^-1.
- So, the energy required to heat 8.0g of water by 1°C is (4.18 calories/g°C) * 8.0g * 100°C = 3344 calories.
4. Calculate the energy required to vaporize the water at its boiling point:
- The heat of vaporization (or latent heat) of water is 540 calories/g.
- So, the energy required to vaporize 8.0g of water is (540 calories/g) * 8.0g = 4320 calories.
5. Finally, sum up all the energy values calculated in steps 1 to 4:
- Total energy required = (16.72 calories) + (637.6 calories) + (3344 calories) + (4320 calories) = 8320.32 calories.
Therefore, it would take approximately 8320.32 calories to convert 8.0g of ice at 0°C to steam at 100°C.