Answer:
We need 1136.4 kJ energy
Explanation:
Step 1: Data given
The mass of water = 369 grams
the initial temperature = -10°C
The finaltemperature = 125 °C
Cm (ice)=36.57 J/(mol⋅∘C)
Cm (water)=75.40 J/(mol⋅∘C)
Cm (steam)=36.04 J/(mol⋅∘C)
ΔHfus=+6010 J/mol
ΔHvap=+40670 J/mol
Step 2: :Calculate the energy needed to heat ice from -10 °C to 0°C
Q = n*C*ΔT
⇒Q = the energy needed to heat ice to 0°C
⇒with n = the moles of ice = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of ice = 36.57 / J/mol°C
⇒ ΔT = the change of temperature = 10°C
Q = 20.48 moles * 36.57 J/mol°C * 10°C
Q = 7489.5 J = 7.490 kJ
Step 3: calculate the energy needed to melt ice to water at 0°C
Q = n* ΔHfus
Q = 20.48 moles * 6010 J/mol
Q = 123084.8 J = 123.08 kJ
Step 4: Calculate energy needed to heat water from 0°C to 100 °C
Q = n*C*ΔT
⇒Q = the energy needed to heat water from 0°C to 100 °C
⇒with n = the moles of water = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of water =75.40 J/(mol⋅∘C)
⇒ ΔT = the change of temperature = 100°C
Q = 20.48 moles * 75.40 J/mol°C* 100°C
Q = 154419.2 J = 154.419 kJ
Step 5: Calculate energy needed to vapourize water to steam at 100°C
Q = 20.48 moles * 40670 J/mol
Q = 832921.6 J = 832.922 kJ
Step 6: Calculate the energy to heat steam from 100 °C to 125 °C
⇒Q = the energy needed to heat steam from 100 °C to 125 °C
⇒with n = the moles of steam = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of steam = 36.04 J/(mol⋅∘C)
⇒ ΔT = the change of temperature = 25 °C
Q = 20.48 moles * 36.04 J/mol*°C * 25°C
Q = 18452.5 J = 18.453 kJ
Step 7: Calculate total energy needed
Q = 7489.5 J + 123084.8 J + 154419.2 J + 832921.6 J + 18452.5 J
Q = 1136367.6 J
Q = 1136.4 kJ
We need 1136.4 kJ energy
