Calculate the amount of energy in kilojoules needed to change 369 gg of water ice at - -10 ∘C∘C to steam at 125 ∘C∘C. The following constants may be useful: Cm (ice) = 36.57 J / (mol⋅∘C) Cm (ice) = 36.57 J / (mol⋅∘C) Cm (water) = 75.40 J / (mol⋅∘C) Cm (water) = 75.40 J / (mol⋅∘C) Cm (steam) = 36.04 J / (mol⋅∘C) Cm (steam) = 36.04 J / (mol⋅∘C) ΔHfus=+6.01 kJ/molΔHfus=+6.01 kJ/mol ΔHvap=+40.67 kJ/molΔHvap=+40.67 kJ/mol Express your answer with the appropriate units.

Question

Chemistry

Alan Simpson

14 September, 20:07

Calculate the amount of energy in kilojoules needed to change 369 gg of water ice at - -10 ∘C∘C to steam at 125 ∘C∘C. The following constants may be useful: Cm (ice) = 36.57 J / (mol⋅∘C) Cm (ice) = 36.57 J / (mol⋅∘C) Cm (water) = 75.40 J / (mol⋅∘C) Cm (water) = 75.40 J / (mol⋅∘C) Cm (steam) = 36.04 J / (mol⋅∘C) Cm (steam) = 36.04 J / (mol⋅∘C) ΔHfus=+6.01 kJ/molΔHfus=+6.01 kJ/mol ΔHvap=+40.67 kJ/molΔHvap=+40.67 kJ/mol Express your answer with the appropriate units.

+3

Answers (1)

Know the Answer?

Kingston Mayer · Answer 1

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