how much heat is needed to change 20g of ice at negative 10 degrees c to water vapor at 200 degrees c
add the following energyies:
heat to heat ice at -10 to 0
heat to melt 20 g ice at 0C
Heat to heat water from 0C to 100C
heat to vaporize 20gwater at 100C.
heat to heat steam from 100C to 200C
To determine how much heat is needed to change 20g of ice at -10 degrees Celsius to water vapor at 200 degrees Celsius, we need to consider the different phases of water and the specific heat capacities involved. The process involves three different steps:
1. Heating the ice from -10 degrees Celsius to 0 degrees Celsius (Phase transition from solid to liquid).
2. Heating the liquid water from 0 degrees Celsius to 100 degrees Celsius.
3. Heating the water vapor from 100 degrees Celsius to 200 degrees Celsius.
Let's break down each step and calculate the amount of heat required:
1. Heating the ice from -10 degrees Celsius to 0 degrees Celsius:
The specific heat of ice is approximately 2.09 J/g°C. Thus, the heat required to raise the temperature of 20g of ice from -10°C to 0°C can be calculated as follows:
Heat = Mass * Specific heat * Temperature change
Heat = 20g * 2.09 J/g°C * (0°C - (-10°C))
Heat = 20g * 2.09 J/g°C * 10°C
Heat = 418 J
2. Melting the ice at 0 degrees Celsius to liquid water:
The heat of fusion for ice is approximately 333.55 J/g. Therefore, the heat required to melt the ice at 0°C can be calculated as follows:
Heat = Mass * Heat of fusion
Heat = 20g * 333.55 J/g
Heat = 6671 J
3. Heating the liquid water from 0 degrees Celsius to 100 degrees Celsius:
The specific heat of liquid water is approximately 4.18 J/g°C. We need to calculate the heat required to raise the temperature of 20g of water from 0°C to 100°C:
Heat = Mass * Specific heat * Temperature change
Heat = 20g * 4.18 J/g°C * (100°C - 0°C)
Heat = 20g * 4.18 J/g°C * 100°C
Heat = 8360 J
4. Heating the water vapor from 100 degrees Celsius to 200 degrees Celsius:
The specific heat of water vapor is approximately 2.02 J/g°C. The heat required to raise the temperature of 20g of water vapor from 100°C to 200°C can be calculated as:
Heat = Mass * Specific heat * Temperature change
Heat = 20g * 2.02 J/g°C * (200°C - 100°C)
Heat = 20g * 2.02 J/g°C * 100°C
Heat = 4040 J
Now, we will add up the heat required for each step:
Total heat = Heat of step 1 + Heat of step 2 + Heat of step 3 + Heat of step 4
Total heat = 418 J + 6671 J + 8360 J + 4040 J
Total heat = 19489 J
Therefore, approximately 19,489 Joules of heat are needed to change 20g of ice at -10 degrees Celsius to water vapor at 200 degrees Celsius.