How much heat i released to the environment as 245 g of steam at 140°C is cooled to -15°C?

1 Answer
Dec 14, 2016

Answer:

Warning! Long answer! The process releases 765 kJ of heat into the environment.

Explanation:

There are five separate heats involved in this problem:

  • #q_1# = heat required to cool the steam from 140 °C to 100 °C
  • #q_2# = heat required to condense the steam to water at 100 °C
  • #q_3# = heat required to cool the water from 100 °C to 0 °C
  • #q_4# = heat required to freeze the water to ice at 0 °C
  • #q_5# = heat required to cool the ice from 0 °C to -15 °C

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

#= mc_1ΔT_1 + mΔ_text(cond)H + mc_3ΔT_3 + mΔ_text(fus)H + mc_5ΔT_5#

where

#q_1, q_2, q_3, q_4,# and #q_5# are the heats involved in each step

#m# is the mass of the sample

#ΔT = T_"f" -T_"i"#

#c_1 = "the specific heat capacity of steam" = "2.010 J·°C"^"-1""g"^"-1"#

#c_3 = "the specific heat capacity of water" = "4.184 J·°C"^"-1""g"^"-1"#

#c_5 = "the specific heat capacity of ice" = "2.010 J·°C"^"-1""g"^"-1"#

#Δ_text(cond)H = "the enthalpy of condensation of steam" = "-2258 J·g"^"-1"#

#Δ_text(sol)H = "the enthalpy of solidification of water" = "-333.55 J·g"^"-1"#

#bbq_1#

#ΔT_1 = "100 °C - 140°C" = "-40 °C"#

#q_1 = mc_1ΔT_1 = 245 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-40") color(red)(cancel(color(black)("°C"))) = "-19 700 J"#

#bbq_2#

#q_2 = 245 color(red)(cancel(color(black)("g"))) × ("-2258")color(white)(l) "J"·color(red)(cancel(color(black)("g"^"-1"))) = "-553 200 kJ"#

#bbq_3#

#ΔT = "0 °C - 100 °C" = "-100 °C"#

#q_3 = mcΔT = 245 color(red)(cancel(color(black)("g"))) × 4.184 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-100") color(red)(cancel(color(black)("°C"))) = "-102 500 J"#

#bbq_4#

#q_4 = 245 color(red)(cancel(color(black)("g"))) × ("-333.55")color(white)(l) "J"·color(red)(cancel(color(black)("g"^"-1"))) = "-81 720 kJ"#

#bbq_5#

#ΔT_5 = "-15 °C - 0 °C" = "-15 °C"#

#q_5 = mcΔT = 245 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-15") color(red)(cancel(color(black)("°C"))) = "-7390 J"#

#q = q_1 + q_2 + q_3 + q_4 + q_5#
#= "-19 700 J" - "553 200 J" - "102 500 J" - "81 720 J" - "7390 J" = "-765 000 J" = "-765 kJ"#

The process releases 765 kJ of heat into the environment.