# Question fb65f

Dec 21, 2017

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

#### Explanation:

Here you have various processes:
First steam cools down to ${100}^{\circ}$; then it condenses into water and then water is cooled down to ${40}^{\circ}$ releasing a total amount of Energy/Heat, ${Q}_{\text{tot}}$:

We can change into grams $2 k g = 2000 g$ and write:

${Q}_{\text{tot}} = {Q}_{1} + {Q}_{2} + {Q}_{3}$

${Q}_{\text{tot}} = 2000 \cdot 2 | 100 - 174 | + 2000 \cdot 2258 + 2000 \cdot 4.18 | 40 - 100 | = 5 , 313 , 600 J$

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

You should get 5320 kJ.

#### Explanation:

Here is a schematic cooling curve for water.

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

$m \textcolor{w h i t e}{m m m} = \text{the mass of the sample}$

${C}_{\textrm{s t e a m}} \textcolor{w h i t e}{m} = \text{the specific heat capacity of steam}$

${C}_{\textrm{w a t e r}} \textcolor{w h i t e}{m} = \text{the specific heat capacity of water}$

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

ΔTcolor(white)(mm) = T_"f" -T_"i"

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

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"

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

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"

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

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"#

$q = {q}_{1} + {q}_{2} + {q}_{3} = \text{(-298 - 4520 - 510) kJ" = "-5320 kJ}$

The process releases 5320 kJ of heat energy.