Question #fb65f

2 Answers
Dec 21, 2017

Answer:

I tried this but check my maths anyway....

Explanation:

Here you have various processes:
First steam cools down to #100^@#; then it condenses into water and then water is cooled down to #40^@# releasing a total amount of Energy/Heat, #Q_"tot"#:
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We can change into grams #2kg=2000g# and write:

#Q_"tot"=Q_1+Q_2+Q_3#

#Q_"tot"=2000*2|100-174|+2000*2258+2000*4.18|40-100|=5,313,600J#

I used the absolute value to avoid confusion; it is heat that goes out of the system and is considered conventionally negative (heat in=positive, heat out=negative) but I didn't want to confuse you so remember that it is heat released into the environment.

Feb 16, 2018

Answer:

You should get 5320 kJ.

Explanation:

Here is a schematic cooling curve for water.

Cooling curve

You are starting with the vapour at the top of the red graph, cooling it to 100 °C, condensing the steam along the horizontal line, and then cooling the water to the bottom of the red curve.

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

  • #q_1# = heat removed by cooling the steam from 174 °C to 100 °C
  • #q_2# = heat removed by condensing the steam to water at 100 °C
  • #q_3# = heat removed by cooling the water from 100 °C to 40 °C

#q = q_1 + q_2 + q_3 = mC_text(steam)ΔT_1 + nΔ_text(cond)H + mC_text(waterl)ΔT_3#

where

#mcolor(white)(mmm) = "the mass of the sample"#

#C_text(steam)color(white)(m) = "the specific heat capacity of steam"#

#C_text(water)color(white)(m) = "the specific heat capacity of water"#

#Δ_text(cond)H = "the enthalpy of condensation of steam"#

#ΔTcolor(white)(mm) = T_"f" -T_"i"#

#bbq_1#

#ΔT_1 = "100 °C - 174 °C = -74 °C"#

#q_1 = mC_text(steam)ΔT = 2000 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-74" color(red)(cancel(color(black)("°C")))) = "-298 000 J" = "-298 kJ"#

#bbq_2#

#q_2 = mΔ_text(cond)H = 2000 color(red)(cancel(color(black)("g"))) × ("-2260 J"·color(red)(cancel(color(black)("g"^"-1")))) = "-4 520 000 J = -4520 kJ"#

#bbq_3#

#ΔT_3 = "40 °C - 100 °C = -60 °C"#

#q_1 = mC_text(water)ΔT = 2000 color(red)(cancel(color(black)("g"))) × 4.184 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-60" color(red)(cancel(color(black)("°C")))) = "-510 000 J" = "-510 kJ"#

#q = q_1 + q_2 + q_3 = "(-298 - 4520 - 510) kJ" = "-5320 kJ"#

The process releases 5320 kJ of heat energy.