# When 150 g of ammonium nitrate dissolves in 450 g of water, the temperature of the solution decreases from 22 °C to 6.9 °C. What is the enthalpy of solution in joules per gram of ammonium nitrate?

## Assume that the specific heat capacity of the solution is the same as that of pure water.

Jul 21, 2017

The enthalpy of solution is 250 J/g.

#### Explanation:

There are two heat transfers involved in this problem.

$\text{heat of solution of NH"_4"NO"_3 + "heat lost by water} = 0$
$\textcolor{w h i t e}{m m m m m m m} {q}_{1} \textcolor{w h i t e}{m m m m m l} + \textcolor{w h i t e}{m m m m} {q}_{2} \textcolor{w h i t e}{m m m m} = 0$
color(white)(mmmmml)m_1Δ_text(sol)Hcolor(white)(mmmll) + color(white)(mmll)m_2CΔT color(white)(mm)= 0

Let's calculate the heats separately.

${m}_{1} = \text{150 g}$

q_1 = m_1Δ_text(sol)H = 150Δ_text(sol)Hcolor(white)(l) "g"

${m}_{2} = \text{(150 + 450) g" = "600 g}$
$C = \text{4.18 J·g·°C"^"-1}$
ΔT = "(6.9 - 22) °C" = "-15.1 °C"

q_2 = m_2CΔT = 600 color(red)(cancel(color(black)("g"))) × "4.184J"·color(red)(cancel(color(black)("g"^"-1"·"°C"^"-1"))) × ("-15.1") color(red)(cancel(color(black)("°C"))) = "-37 900 J"

q_1 + q_2 = 150Δ_text(sol)Hcolor(white)(l) "g" - "37 870 J"" = 0

Δ_text(sol)H = "37 900 J"/"150 g" = "250 J/g"

Note: The answer can have only two significant figures, because that is we have for the value of ΔT.