Calculate the number of joules given off when 32.0 grams of steam cools from 110.0 °C to ice at -40.0 °C.

Answer 1

The amount of energy given off is 99 600 J.

There are five heats to consider:

#q_1# = heat lost on cooling steam from 110.0 °C to 100 °C.

#q_2# = heat lost on condensing steam to water at 100 °C.

#q_3# = heat lost on cooling water from 100 °C to 0°C.

#q_4# = heat lost on freezing water to ice at 0 °C.

#q_5# = heat lost on cooling ice from 0 °C to -40.0 °C.

The total heat evolved is

#q = q_1 + q_2 + q_3 + q_4 + q_5#

1. Cooling the Steam

# m = "32.0 g H"_2"O"#

For steam, the specific heat capacity, #c = "2.010 J·g"^"-1""°C"^"-1"#.

#ΔT# = #T_2 – T_1 = "(100.0 - 110.0) °C" = "-10.0 °C"#

#q_1 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "2.010 J·"color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-10.0" color(red)(cancel(color(black)("°C")))) = "-643 J"#

2. Condensing the Steam

#"Heat of condensation = -Heat of vaporization"#

#Δ H_"cond" = ""-ΔH_"vap" = "-2260 J·g"^"-1"#

#q_2 = m Δ H_"cond" = 32.0 color(red)(cancel(color(black)("g")))×("-2260 J·"color(red)(cancel(color(black)("g"^"-1")))) = "-72 320 J"#

3. Cooling the Water

For liquid water, the specific heat capacity, #c = "4.184 J·°C"^"-1""g"^"-1"#.

#ΔT = T_2 – T_1 = "(0 - 100) °C" = "-100 °C"#

#q_3 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "4.184 J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1")))× ("-100"color(red)(cancel(color(black)("°C")))) = "-13 389 J"#

4. Freezing the Water

#"Heat of freezing = -Heat of fusion"#

#"-"ΔH_"fus" = "334 J·g"^"-1"#

#ΔH_"freeze" = "-"ΔH_"fus" = "-334 J·g"^-1"#

#q_4 = "-"m Δ H_"fus" = 32.0 color(red)(cancel(color(black)("g"))) × "-334 J·"color(red)(cancel(color(black)("g"^"-1"))) = "-10 689 J"#

5. Cooling the Ice

The specific heat capacity of ice, #c = "2.03 J·°C"^"-1""g"^"-1"#

#ΔT = T_2 – T_1 = "(-40.0 - 0) °C" = "-40.0 °C"#

#q_5 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × 2.03 "J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × (color(red)(cancel(color(black)("-40.0 °C")))) = "-2598 J"#

Adding them all up

#q = q_1 + q_2 + q_3 + q_4 + q_5 = "(-643 – 72 320 – 13 389 – 10 689 - 2598) J" = "-99 600 J"#

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To calculate the energy change, use the formula:

Q = mcΔT

Where Q is the heat energy, m is the mass, c is the specific heat, and ΔT is the change in temperature.

For steam to ice transition, you need to consider the heat for cooling steam to water and then freezing water to ice. Use the specific heat values for water and steam, and the heat of fusion for water.

Calculate Q1 for cooling steam to water: Q1 = (32.0 g) * (2.0 J/g°C) * (110.0°C - 100.0°C)

Calculate Q2 for phase change (steam to water): Q2 = (32.0 g) * (2260 J/g)

Calculate Q3 for cooling water to ice: Q3 = (32.0 g) * (4.18 J/g°C) * (-40.0°C - 0.0°C)

Add Q1, Q2, and Q3 to get the total heat energy.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7