**247 kJ**

*Note on significant figures:*

*I've assumed the freezing point and boiling point temperatures to be exact.*

*Also, I kept an extra sig fig in the answer for each step.*

*Source of constants:*

What is the answer in kJ?

Tutors, sign in to answer this question.

David L. | Chemistry TutorChemistry Tutor

This question requires five separate calculations: (1) energy needed to warm ice at −24.0ºC to its melting point at 0ºC, (2) energy needed to melt ice at OºC, (3) energy needed to warm liquid water from 0ºC to its boiling point at 100ºC, (4) energy needed to boil water at 100ºC, and (5) energy needed to warm steam from 100ºC to 169ºC. The sum of these five is the answer.

For the two phase changes, the enthalpy of fusion (333.55 kJ/kg) and enthalpy of vaporization (2257 kJ/kg) are needed, and the heat required is q = m • ΔH

For the three warming steps, specific heat capacities are needed for each step: ice is 2.05 J/g•ºC, liquid water is 4.184 J/g•ºC, and steam is 2.080 J/g•ºC, and the heat required is q = m • C • ΔT

Step 1: Energy needed for warming the ice from −24.0ºC to 0ºC. ΔT = 24.0ºC

q = m • C • ΔT = 77.0 g x (2.05 J/g•ºC) x (1 kJ / 1000 J) x (24.0 ºC) = 3.788 kJ

Step 2: Energy needed for melting the ice at 0ºC.

q = m • ΔH = 77.0 g x (1 kg/1000 g) x (333.55 kJ/kg) = 25.68 kJ

Step 3: Energy needed for warming the liquid water from 0ºC to 100ºC. ΔT = 100ºC

q = m • C • ΔT = 77.0 g x (4.184 J/g•ºC) x (1 kJ / 1000 J) x (100 ºC) = 32.22 kJ

Step 4: Energy needed for boiling the water at 100ºC.

q = m • ΔH = 77.0 g x (1 kg/1000 g) x (2257 kJ/kg) = 173.8 kJ

Step 5: Energy needed for warming the steam from 100ºC to 169.0ºC. ΔT = 69.0ºC

q = m • C • ΔT = 77.0 g x (2.080 J/g•ºC) x (1 kJ / 1000 J) x (69.0 ºC) = 11.05 kJ

Summing the energies required for each step yields the answer:

3.788 kJ + 25.68 kJ + 32.32 kJ + 173.8 kJ + 11.05 kJ =
**247 kJ**

Morgan,

Step 1 amount of energy required to melt the ice

q = m·ΔHf

where

q = heat energy

m = mass

ΔHf = heat of fusion

q = (77 g)x(334 J/g)

q = 25,718 J

q = m·ΔHf

where

q = heat energy

m = mass

ΔHf = heat of fusion

q = (77 g)x(334 J/g)

q = 25,718 J

Step 2 amount of energy required to vaporize the water

q = m·ΔHv

where

q = heat energy

m = mass

ΔHv = heat of vaporization

q = (77 g)x(2257 J/g)

q = 173,789 J

where

q = heat energy

m = mass

ΔHv = heat of vaporization

q = (77 g)x(2257 J/g)

q = 173,789 J

Adding gives 199,507 J or 199.5 kJ

Jim

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.