# Another chemistry question over thermodynamics?

Jun 13, 2018

${T}_{\textrm{f}} = \text{71 °C}$

#### Explanation:

There are two heat transfers to consider:

$\text{heat released by reaction + heat absorbed by water = 0}$
$\textcolor{w h i t e}{m m m m m m} {q}_{1} \textcolor{w h i t e}{m m m m m l} + \textcolor{w h i t e}{m m m m m m} {q}_{2} \textcolor{w h i t e}{m m m m l} = 0$
color(white)(mmmmm)nΔ _text(r)H color(white)(mmmml)+ color(white)(mmmm)mC_text(s)ΔT color(white)(mmml)= 0

Let's calculate each of these heats separately.

Step 1. Calculate ${q}_{1}$

q_1 = nΔ_text(r)H = 0.110 color(red)(cancel(color(black)("mol"))) × ("-131 kJ"·color(red)(cancel(color(black)("mol"^"-1")))) = "-14.41 kJ" = "-14 410 J"

Step 2. Calculate ${q}_{2}$

$m \textcolor{w h i t e}{l} = \text{75 g}$
${C}_{\textrm{s}} \textcolor{w h i t e}{l} = \text{4.184 J·°C"^"-1""g"^"-1}$
ΔT = ?

q_2 = 75 color(red)(cancel(color(black)("g"))) × "4.184 J·°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × ΔT = 314 color(white)(l)ΔTcolor(white)(l) "J·°C"^"-1"

Step 3. Calculate ${T}_{\textrm{f}}$

ΔT = T_text(f) - T_text(i)

${q}_{1} + {q}_{2} = 0$

$\text{-14 410" color(red)(cancel(color(black)("J"))) + 314 color(white)(l)ΔTcolor(red)(cancel(color(black)("J")))·"°C"^"-1} = 0$

ΔT = "14 410"/("314 °C"^"-1") = "45.9 °C"

T_text(f) = T_text(i) + ΔT = "(25 + 45.9) °C = 71 °C"