# Question 85b26

Apr 1, 2017

The temperature change would be -6.68 °C.

#### Explanation:

This is a disguised calorimetry problem, with two heat transfers involved.

$\textcolor{w h i t e}{m m m m} \text{NH"_4"NO"_3"(s)" color(white)(mmml)→ color(white)(m)"NH"_4"NO"_3"(aq)}$

$\text{heat of solution of NH"_4"NO"_3 + "heat lost by water} = 0$

$\textcolor{w h i t e}{m m m m m l} {q}_{1} \textcolor{w h i t e}{m m m m m m m} + \textcolor{w h i t e}{m m m m} {q}_{2} \textcolor{w h i t e}{m m m l l} = 0$

color(white)(mmmll)nΔ_text(soln)Hcolor(white)(mmmmmll) + color(white)(mmll)mCΔT = 0

In this problem,

n = 500 color(red)(cancel(color(black)("g NH"_4"NO"_3))) × (1 "mol NH"_4"NO"_3)/(80.04 color(red)(cancel(color(black)("g NH"_4"NO"_3)))) = "6.247 mol NH"_4"NO"_3"

Δ_text(soln)H = "25.41 kJ· mol"^"−1"

q_1 = 6.247 color(red)(cancel(color(black)("mol"))) × "25 410 J"·color(red)(cancel(color(black)("mol"^"-1"))) = "158 700 J"

I am guessing that you are using US gallons. Then

m = 1.50 color(red)(cancel(color(black)("gal"))) × (3.785 color(red)(cancel(color(black)("L"))))/(1 color(red)(cancel(color(black)("gal")))) × ("1000 g")/(1 color(red)(cancel(color(black)("L")))) = "5678 g"

$C = \text{4.184 J·g"^"-1""°C"^"-1}$

 q_2 = 5678 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g"^"-1")))"°C"^"-1" × ΔT = "23 750 J·°C"^"-1"·ΔT

q_1 + q_2 = "158 700" color(red)(cancel(color(black)("J"))) + "23 750" color(red)(cancel(color(black)("J")))"°C"^"-1"·ΔT = 0

ΔT = "-158 700"/"23 750 °C"^"-1" = "-6.68 °C"#