Question #05bc9

1 Answer
Mar 28, 2017

Answer:

WARNING! Long answer! You need to add 91.0 kcal of heat.

Explanation:

A typical heating curve of water is shown below.

figures.boundless-cdn.com

There are five separate heats involved in this problem:

  • #q_1# = heat required to warm the ice from -15 °C to 0 °C
  • #q_2# = heat required to melt the ice to water at 0 °C
  • #q_3# = heat required to warm the water from 0 °C to 100 °C
  • #q_4# = heat required to boil the water to ice at 1000 °C
  • #q_5# = heat required to heat the steam from 100 °C to -105.0 °C

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

#= mc_1ΔT_1 + mΔ_text(fus)H + mc_3ΔT_3 + mΔ_text(vap)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_1color(white)(mm) = "the specific heat capacity of ice" = "0.50 cal·°C"^"-1""g"^"-1"#

#c_3color(white)(mm) = "the specific heat capacity of water" = "1.00 cal·°C"^"-1""g"^"-1"#

#c_5 color(white)(mm)= "the specific heat capacity of steam" = "0.48 cal·°C"^"-1""g"^"-1"#

#Δ_text(fus)H = "the enthalpy of fusion of ice" = "79.5 cal·g"^"-1"#

#Δ_text(vap)H = "the enthalpy of vaporization of water" = "539 cal·g"^"-1"#

#bbq_1#

#ΔT_1 = "0 °C - (-15 °C)" = "15 °C"#

#q_1 = mc_1ΔT_1 = 125.00 color(red)(cancel(color(black)("g"))) × 0.50 color(white)(l)"cal"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 15 color(red)(cancel(color(black)("°C"))) = "938 cal"#

#bbq_2#

#q_2 = 125.00 color(red)(cancel(color(black)("g"))) × 79.5color(white)(l) "cal"·color(red)(cancel(color(black)("g"^"-1"))) = "9938 cal"#

#bbq_3#

#ΔT = "100 °C - 0 °C" = "100 °C"#

#q_3 = mcΔT = 125.00 color(red)(cancel(color(black)("g"))) × 1.00 color(white)(l)"cal"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 100 color(red)(cancel(color(black)("°C"))) = "12 500 cal"#

#bbq_4#

#q_4 = 125.00 color(red)(cancel(color(black)("g"))) × 539color(white)(l) "cal"·color(red)(cancel(color(black)("g"^"-1"))) = "67 380 cal"#

#bbq_5#

#ΔT_5 = "105.0 °C - 1000 °C" = "5.0 °C"#

#q_5 = mcΔT = 125.00 color(red)(cancel(color(black)("g"))) × 0.48 color(white)(l)"cal"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 5.0 color(red)(cancel(color(black)("°C"))) = "300 cal"#

#q = q_1 + q_2 + q_3 + q_4 + q_5 = "938 cal" + "9938 cal" + "12 500 cal" + "67 380 cal" + "300 cal" = "91 000 cal" = "91.0 kcal"#

The process releases 91.0 kcal of heat.