Question #64f37

1 Answer
Sep 5, 2016

Answer:

The amount of heat required is 12.68 kJ.

Explanation:

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There are four separate heats involved in this problem:

  1. #q_1# = heat required to warm the ice from -11.09 °C to 0 °C
  2. #q_2# = heat required to melt the ice to water at 0 °C
  3. #q_3# = heat required to warm the water from 0 °C to 100 °C
  4. #q_4# = heat required to convert the water to steam at 100 °C

#q = q_1 + q_2 + q_3 + q_4 = nc_1ΔT_1 +nΔ_text(fus)H + nc_2ΔT_2 + nΔ_text(vap)H#

#bbq_1#

#n = "0.232 mol"#
#c_1 = "37.6 J·°C"^"-1""mol"^"-1"#
#ΔT = "0.0 °C - (-11.09 °C)" = "11.09 °C"#

#q_1 = ncΔT = 0.232 color(red)(cancel(color(black)("mol"))) × 37.6 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""mol"^"-1"))) × 11.09 color(red)(cancel(color(black)("°C"))) = "96.74 J" = "0.096 74 kJ"#

#bbq_2#

#Δ_"fus"H = "40.7 kJ·mol"^"-1"#

#q_2 = 0.232 color(red)(cancel(color(black)("mol"))) × 6.02color(white)(l) "kJ"·color(red)(cancel(color(black)("mol"^"-1"))) = "1.397 kJ"#

#bbq_3#

#c_2 = "75.2 J·°C"^"-1""mol""-1"#
#ΔT_2 = "100 °C - 0°C" = "100 °C"#

#q_3 = ncΔT = 0.232 color(red)(cancel(color(black)(" mol"))) × 75.2 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""mol"^"-1"))) × 100 color(red)(cancel(color(black)("°C"))) = "1745 J" = "1.745 kJ"#

#bbq_4#

#Δ_"vap"H = "40.7 J·K"^"-1""mol""-1"#

#q_4 = nΔ_"vap"H = 0.232 color(red)(cancel(color(black)("mol"))) × 40.7color(white)(l) "kJ"·color(red)(cancel(color(black)("mol"^"-1"))) = "9.442 kJ"#

#q = q_1 + q_2 + q_3 + q_4 = "0.096 74 kJ" + "1.397 kJ" + "1.745 kJ" + "9.442 kJ" = "12.68 kJ"#