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

1 Answer

Answer:

The amount of energy given off is 99 600 J.

Explanation:

There are five heats to consider:

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#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"#