Question

How much heat, in kilojoules, must be lost by 155 grams of steam at 125 degrees celsius to convert steam to water and then cool the water to 20.0 degree celsius?

( Note: you must consider that two states of matter are involved im this question and that the temperature listed here are in Celsius: The molar heat of fusion of water is 6.01 kl/mol; The molar hear of vapoization of water is 40.7 kj/mol )

I know the answer is around 410. kj but I get mixed up with the conversions :( someone please help me!! I have a test coming up really soon and I dont understand these type of questions.

Answer 1

Warning! Long Answer. The system must lose 410. kJ.

Explanation:

Here is a schematic cooling curve for water.

You are starting with steam at the top of the red graph, cooling it to its condensation point, condensing the steam to water along the horizontal line, and then cooling the water to the bottom of the red graph.

Thus, there are three separate heat removals involved in this problem:

• `q_1` = heat removed by cooling the steam from 125 °C to 100 °C
• `q_2` = heat removed by condensing the steam at 100 °C
• `q_3` = heat removed by cooling the water from 100 °C to 20 °C

`q = q_1 + q_2 + q_3 = mC_1ΔT_1 + nΔ_text(cond)H + mC_3ΔT_3`

where

`mcolor(white)(mmll) = "the mass of the water"`

`C_1color(white)(mml) =`the specific heat capacity of steam (`"2.010 J·°C"^"-1""g"^"-1"`)

`C_3color(white)(mml) =`the specific heat capacity of water (`"4.184 J·°C"^"-1""g"^"-1"`)

`ΔT color(white)(mm)= T_"f" -T_"i"`

`" ncolor(white)(mmm) =` the moles of water

`Δ_text(cond)H = "-"Δ_text(vap)H =` the molar heat of condensation of water

`bbq_1`

`ΔT_1 = "100 °C - 125 °C = -25 °C"`

`q_1 = mC_1ΔT = 155 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-25" color(red)(cancel(color(black)("°C")))) = "-7790 J" = "-7.79 kJ"`

`bbq_2`

`n = 155 color(red)(cancel(color(black)("g"))) × "1 mol"/(18.02 color(red)(cancel(color(black)("g")))) = "8.602 mol"`

`q_2 = nΔ_text(cond)H = 8.602 color(red)(cancel(color(black)("mol"))) × ("-40.7 kJ"·color(red)(cancel(color(black)("mol"^"-1")))) = "-350.1 kJ"`

`bbq_3`

`ΔT_3 = "20 °C - 100 °C = -80 °C"`

`q_3 = mC_3ΔT = 155 color(red)(cancel(color(black)("g"))) × 4.184 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-80" color(red)(cancel(color(black)("°C")))) = "-51 900 J" = "-51.9 kJ"`

`q = q_1 + q_2 + q_3 = "(-7.79 - 350.1 - 51.9) kJ" = "-410. kJ"`

The system must lose 410. kJ.