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

Dec 14, 2016

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"

${c}_{1} = \text{the specific heat capacity of steam" = "2.010 J·°C"^"-1""g"^"-1}$

${c}_{3} = \text{the specific heat capacity of water" = "4.184 J·°C"^"-1""g"^"-1}$

${c}_{5} = \text{the specific heat capacity of ice" = "2.010 J·°C"^"-1""g"^"-1}$

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"

${\boldsymbol{q}}_{1}$

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"

${\boldsymbol{q}}_{2}$

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

${\boldsymbol{q}}_{3}$

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"

${\boldsymbol{q}}_{4}$

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"

${\boldsymbol{q}}_{5}$

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"

$q = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5}$
$= \text{-19 700 J" - "553 200 J" - "102 500 J" - "81 720 J" - "7390 J" = "-765 000 J" = "-765 kJ}$

The process releases 765 kJ of heat into the environment.