Chemistry · General chemistry II · Worked example

Find the temperature change of a cold pack

$10.0 g$ of ammonium nitrate, ${NH}_{4} {NO}_{3}$ , dissolves in $100.0 g$ of water at $22.0 ° C$ . Take $Δ H (soln) = + 25.7 kJ / mol$ and $c = 4.184 J / (g \cdot ° C)$ for the solution, and assume no heat is lost. What is the final temperature?

q = n Δ H_{soln}, q_{soln} = m c Δ T

Moles of solute

The $NaCl$ example found $Δ H (soln)$ from a cycle. This one starts from a known $Δ H (soln)$ and asks what the water feels. $Δ H (soln)$ is per mole, so convert the salt first. ${NH}_{4} {NO}_{3}$ is $2 (14.01)$ $+ 4 (1.008)$ $+ 3 (16.00)$ $= 80.05 g / mol$ .

10.0 g \times \frac{1 mol}{80.05 g} = 0.1249 mol

Write the heat absorbed

The heat taken in by the dissolving is the moles times the enthalpy per mole.

q = n Δ H_{soln}

Name the moles

Use the moles from the first step.

n = 0.1249 mol

Name the enthalpy of solution

The value is positive, so the dissolving absorbs heat.

Δ H_{soln} = 25.7 kJ/mol

Substitute

$0.1249 \times 25.7 = 3.21 kJ$ absorbed.

q = (0.1249 mol) (25.7 kJ/mol) = 3.21 kJ

\begin{array}{c} q = (0.1249 mol) (25.7 kJ/mol) \\ = 3.21 kJ \end{array}

Turn to the solution

The only source of that heat is the solution itself, so the solution loses it. Its heat has the opposite sign, written in joules to match $c$ .

q_{soln} = - 3.21 \times 10^{3} J

Write the calorimetry relation

A heat and a temperature change are linked by the mass and the specific heat.

q_{soln} = m c Δ T

Solve for $Δ T$

Divide both sides by $m c$ before putting in numbers.

Δ T = \frac{q_{soln}}{m c}

Name the mass

All of the solution changes temperature, salt included, so $m = 100.0 g + 10.0 g$ $= 110.0 g$ .

m = 110.0 g

Name the specific heat

A dilute water solution is close to water’s value.

c = 4.184 J g^{- 1} \circ C^{- 1}

\begin{array}{c} c = 4.184 \\ J g^{- 1} \circ C^{- 1} \end{array}

Substitute

Joules cancel, and so do grams, leaving $° C$ . $- 3210 \div 460.2 = - 6.97$ . That comes from the rounded $3.21 kJ$ . Carried unrounded, from $0.12492 mol$ , the heat is $3.2105 kJ$ and the drop is $6.98 ° C$ , and either way the solution ends at $15.0 ° C$ .

Δ T = \frac{- 3.21 \times 10^{3} J}{(110.0 g) (4.184 J g^{- 1} \circ C^{- 1})} = - 6.97 \circ C

\begin{array}{c} Δ T = \frac{- 3.21 \times 10^{3}}{(110.0) (4.184)} \circ C \\ = - 6.97 \circ C \end{array}

Find the final temperature

Add the change to the starting temperature. The solution ends near $15.0 ° C$ , colder, as a positive $Δ H (soln)$ predicts.

22.0 \circ C + (- 6.97 \circ C) = 15.0 \circ C

\begin{array}{c} 22.0 \circ C + (- 6.97 \circ C) \\ = 15.0 \circ C \end{array}

Result

The solution cools by $6.97 ° C$ , to about $15.0 ° C$ .

Your turn

1. One like the example

$8.00 g$ of ${NH}_{4} {NO}_{3}$ $(Δ H (soln) = + 25.7 kJ / mol)$ dissolves in $100.0 g$ of water at $25.0 ° C$ . With $c = 4.184 J / (g \cdot ° C)$ and no heat lost, what is the final temperature?

Show the answer and explanation

About $19.3 ° C$ .

$8.00 g \div 80.05 g / mol$ $= 0.09994 mol$ , which absorbs $0.09994 \times 25.7$ $= 2.568 kJ$ . The $108.0 g$ of solution loses $2568 J$ , so $Δ T = - 2568 \div (108.0 \times 4.184)$ $= - 5.68 ° C$ . $25.0 - 5.68 = 19.3 ° C$ .

\begin{array}{l} 8.00 g {NH}_{4} {NO}_{3} \times \frac{1 mol {NH}_{4} {NO}_{3}}{80.05 g {NH}_{4} {NO}_{3}} = 0.09994 mol {NH}_{4} {NO}_{3} \\ (0.09994 mol) (25.7 kJ/mol) = 2.568 kJ \\ \frac{- 2568 J}{(108.0 g) (4.184 J g^{- 1} \circ C^{- 1})} = - 5.68^{\circ} C \\ 25.0^{\circ} C + (- 5.68^{\circ} C) = 19.3^{\circ} C \end{array}

\begin{array}{c} 8.00 g {NH}_{4} {NO}_{3} \\ \times \frac{1 mol}{80.05 g} \\ = 0.09994 mol {NH}_{4} {NO}_{3} \\ (0.09994 mol) (25.7 kJ/mol) \\ = 2.568 kJ \\ \frac{- 2568 J}{(108.0 g) (4.184 J g^{- 1} \circ C^{- 1})} \\ = - 5.68^{\circ} C \\ 25.0^{\circ} C + (- 5.68^{\circ} C) \\ = 19.3^{\circ} C \end{array}

2. A variation

$4.00 g$ of sodium hydroxide, $NaOH$ $(Δ H (soln) = - 44.5 kJ / mol)$ , dissolves in $100.0 g$ of water at $20.0 ° C$ . With $c = 4.184 J / (g \cdot ° C)$ and no heat lost, what is the final temperature?

Show the answer and explanation

About $30.2 ° C$ .

Now $Δ H (soln)$ is negative, so the dissolving releases heat and the solution gains it. $4.00 g \div 40.00 g / mol$ $= 0.1000 mol$ , which releases $4.45 kJ$ . The $104.0 g$ of solution gains $+ 4450 J$ , so $Δ T = 4450 \div (104.0 \times 4.184)$ $= + 10.2 ° C$ , and $20.0 + 10.2 = 30.2 ° C$ .

\begin{array}{l} 4.00 g NaOH \times \frac{1 mol NaOH}{40.00 g NaOH} = 0.1000 mol NaOH \\ (0.1000 mol) (- 44.5 kJ/mol) = - 4.45 kJ \\ \frac{4450 J}{(104.0 g) (4.184 J g^{- 1} \circ C^{- 1})} = 10.2^{\circ} C \\ 20.0^{\circ} C + 10.2^{\circ} C = 30.2^{\circ} C \end{array}

\begin{array}{c} 4.00 g NaOH \times \frac{1 mol NaOH}{40.00 g NaOH} \\ = 0.1000 mol NaOH \\ (0.1000 mol) (- 44.5 kJ/mol) \\ = - 4.45 kJ \\ \frac{4450 J}{(104.0 g) (4.184 J g^{- 1} \circ C^{- 1})} \\ = 10.2^{\circ} C \\ 20.0^{\circ} C + 10.2^{\circ} C = 30.2^{\circ} C \end{array}

3. Somewhere new

How much ${NH}_{4} {NO}_{3}$ $(Δ H (soln) = + 25.7 kJ / mol)$ must dissolve to cool $200.0 g$ of water by $10.0 ° C$ ? Count only the water’s mass in $q = m c Δ T$ , take $c = 4.184 J / (g \cdot ° C)$ and assume no heat is lost.

Show the answer and explanation

About $26.1 g$ .

Run the method backward. Cooling $200.0 g$ of water by $10.0 ° C$ removes $(200.0) (4.184) (10.0)$ $= 8368 J$ , or $8.37 kJ$ . That much heat dissolves $8.37 \div 25.7 = 0.326 mol$ of salt, and $0.326 mol \times 80.05 g / mol$ $= 26.1 g$ . The salt’s own mass, ignored here, would add a little to the heat needed.

\begin{array}{l} (200.0 g) (4.184 J g^{- 1} \circ C^{- 1}) (10.0^{\circ} C) = 8.37 \times 10^{3} J \\ \frac{8.37 kJ}{25.7 kJ/mol} = 0.326 mol \\ 0.326 mol {NH}_{4} {NO}_{3} \times \frac{80.05 g {NH}_{4} {NO}_{3}}{1 mol {NH}_{4} {NO}_{3}} = 26.1 g {NH}_{4} {NO}_{3} \end{array}

\begin{array}{c} (200.0 g) (4.184 J g^{- 1} \circ C^{- 1}) \\ \times (10.0^{\circ} C) = 8.37 \times 10^{3} J \\ \frac{8.37 kJ}{25.7 kJ/mol} = 0.326 mol \\ 0.326 mol {NH}_{4} {NO}_{3} \\ \times \frac{80.05 g}{1 mol} \\ = 26.1 g {NH}_{4} {NO}_{3} \end{array}

Keep exploring

Open the steps in a Chemistry box and change the $10.0 g$ to $20.0 g$ of ${NH}_{4} {NO}_{3}$ . The heat doubles to $6.42 kJ$ , but the solution also grows to $120.0 g$ , so the drop is $12.8 ° C$ , a little less than double.

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Moles of solute

Write the heat absorbed

Name the moles

Name the enthalpy of solution

Substitute

Turn to the solution

Write the calorimetry relation

Solve for Δ⁢TΔT

Name the mass

Name the specific heat

Substitute

Find the final temperature

Result

Your turn

1. One like the example

About 19.3 °⁢C19.3 ∘C.

2. A variation

About 30.2 °⁢C30.2 ∘C.

3. Somewhere new

About 26.1 g26.1 g.

Keep exploring

Study it, then try it

Solve for $Δ T$

About $19.3 ° C$ .

About $30.2 ° C$ .

About $26.1 g$ .