Answer: approximately 46,415 Joules of energy
Step-by-step explanation:
Heating the ice to its melting point (0 degrees Celsius):
The specific heat capacity of ice is approximately 2.09 J/g°C.
The temperature change required is 0°C - (-20°C) = 20°C.
The energy required to heat the ice to 0 degrees Celsius is:
Energy = mass × specific heat capacity × temperature change
= 15 g × 2.09 J/g°C × 20°C
= 626 J
Melting the ice at 0 degrees Celsius:
The heat of fusion (enthalpy of fusion) for ice is approximately 334 J/g.
The energy required to melt the ice is:
Energy = mass × heat of fusion
= 15 g × 334 J/g
= 5010 J
Heating the water from 0 degrees Celsius to 100 degrees Celsius:
The specific heat capacity of water is approximately 4.18 J/g°C.
The temperature change required is 100°C - 0°C = 100°C.
The energy required to heat the water is:
Energy = mass × specific heat capacity × temperature change
= 15 g × 4.18 J/g°C × 100°C
= 6270 J
Vaporizing the water at 100 degrees Celsius:
The heat of vaporization (enthalpy of vaporization) for water is approximately 2260 J/g.
The energy required to vaporize the water is:
Energy = mass × heat of vaporization
= 15 g × 2260 J/g
= 33,900 J
Heating the steam from 100 degrees Celsius to 120 degrees Celsius:
The specific heat capacity of steam is approximately 2.03 J/g°C.
The temperature change required is 120°C - 100°C = 20°C.
The energy required to heat the steam is:
Energy = mass × specific heat capacity × temperature change
= 15 g × 2.03 J/g°C × 20°C
= 609 J
The total energy required is the sum of all the individual steps:
Total energy = Energy of step 1 + Energy of step 2 + Energy of step 3 + Energy of step 4 + Energy of step 5
= 626 J + 5010 J + 6270 J + 33,900 J + 609 J
= 46,415 J
Therefore, it would require approximately 46,415 Joules of energy to heat 15 grams of ice at -20 degrees Celsius to 120 degrees Celsius of steam.