Chapter 5, Problem 104AP

A quantity of $\text{85}\text{.0 mL of 0}\text{.600 }M\text{ HCl}$ is mixed with $\text{85}\text{.0 mL of 0}\text{.600 }M\text{ KOH}$ in a constant-pressure calorimeter. The initial temperature of both solutions is the same at $\text{17}\text{.35}°\text{C,}$ and the final temperature of the mixed solution is $\text{19}\text{.02°C,}$ What is the heat capacity of the calorimeter? Assume that the specific heat of the solutions is the same as that of water and the molar heat of neutralization is $\text{-56}\text{.2 kJ/mol}$.

Interpretation Introduction

Interpretation:

The heat capacity of a calorimeter, when the given amount of $\text{0}\text{.600 M HCl}$ and $\text{0}\text{.600 M KOH}$ mixed, is to be calculated.

Concept introduction:

Heat absorbed or released in the reaction is given by:

$q=sm\Delta T$

Here, $s$ is the specific heat, $m$ is the mass of the sample undergoing the temperature change, and $\Delta T$ is the change in temperature.

Specific heat is the heat required to increase the temperature of $1\text{ g}$ of a substance by1°C. Its S.I. unit is $\left(\text{J}/\text{g}{\text{.}}^{\text{o}}\text{C}\right)$.

Heat absorbed or released in the reaction is also given by:

$q=C\Delta T$

Here, $C$ is the heat capacity of a body and $\Delta T$ is change in temperature.

Heat capacity is defined as the heat required to increase the temperature of a body by1°C.

Its S.I. unit is $\left(\text{J}{/}^{\text{o}}\text{C}\right)$.

Concentration (molarity) of a solution is given by:

$M=\frac{n}{V}$

Here, $n$ is the number of moles and $V$ is the volume of the solution.

Answer to Problem 104AP

Solution: The heat capacity of a calorimeter is $1\times {10}^3\text{ kJ}{/}^{\text{o}}\text{C}$.

Explanation of Solution

Given information:

Molar heat of neutralisation is $-56.2\text{ kJ}/\text{mol}$.

$T_i={17.35}^{\circ }\text{C}$

$T_f={19.02}^{\circ }\text{C}$

Specific heat of the solution = Specific heat of water

Concentration (molarity) of a solution is given by:

$M=\frac{n}{V}$

It can be rewritten as

$M\times V=n$

Equation for the ionization of $\text{HCl}$:

$\text{HCl}\to {\text{H}}^{+}+{\text{Cl}}^{-}$

So, it is clear that $M_{\text{HCl}}=M_{{\text{H}}^{+}}$.

The number of moles of ${\text{H}}^{+}$ can be calculated as follows:

$M_{{\text{H}}^{+}}\times V_{{\text{H}}^{+}}=n_{{\text{H}}^{+}}$

Substitute $\text{0}\text{.600 M}$ for $M_{{\text{H}}^{+}}$ and $\text{85 ml}$ for $V_{{\text{H}}^{+}}$

$\begin{array}{c}{n}_{{\text{H}}^{+}}=\text{0}\text{.600 M}\times \text{85 mL}\\ =\frac{\text{0}\text{.600 M}\times \text{85 L}}{1000}\\ =0.0510\text{ mol}\end{array}$

Equation for the ionization of $\text{KOH}$:

$\text{KOH}\to {\text{K}}^{+}+{\text{OH}}^{-}$

So, it is clear that $M_{\text{KOH}}=M_{{\text{OH}}^{-}}$.

The number of moles of ${\text{OH}}^{-}$ can be calculated as follows:

$M_{{\text{OH}}^{-}}\times V_{{\text{OH}}^{-}}=n_{{\text{OH}}^{-}}$

Substitute $\text{0}\text{.600 M}$ for $M_{{\text{OH}}^{-}}$ and $\text{85 ml}$ for $V_{{\text{OH}}^{-}}$

$\begin{array}{c}{n}_{{\text{OH}}^{-}}=\text{0}\text{.600 M}\times \text{85 mL}\\ =\frac{\text{0}\text{.600 M}\times \text{85 L}}{1000}\\ =0.0510\text{ mol}\end{array}$

Now, the heat produced by the reaction is calculated by:

$q=n\times \Delta {\text{H}}^{\circ }$

Substitute $-56.2\text{ kJ}/\text{mol}$ for $\Delta {\text{H}}^{\circ }$ and $0.0510\text{ mol}$ for $\left(n\right)$ in the above equation

$\begin{array}{c}q=0.0510\text{ mol}\times -56.2\text{ kJ}/\text{mol}\\ \text{=}-\text{2}\text{.8662 kJ}\end{array}$

It is known that $q_{\text{rxn}}=-q_{\text{surr}}$.

Therefore, $q_{\text{surr}}=\text{2}\text{.8662 kJ}$.

However, the heat of the surroundings includes the heat of water and the heat of the calorimeter.

So,

$q_{\text{surr}}=s_{{\text{H}}_2\text{O}}{m}_{{}_{{\text{H}}_2\text{O}}}\Delta T+C_{\text{cal}}\Delta T$ …… (1)

Change in temperature is

$\begin{array}{c}\Delta T=\left(T_f-T_i\right)\\ =\left({19.02}^{\circ }\text{C}-{17.35}^{\circ }\text{C}\right)\\ ={1.67}^{\circ }\text{C}\end{array}$

Mass of ${\text{H}}_2\text{O}$ is calculated by:

$m_{{}_{{\text{H}}_2\text{O}}}=V_{{}_{{\text{H}}_2\text{O}}}\times d_{{}_{{\text{H}}_2\text{O}}}+n_{{}_{{\text{H}}_2\text{O}}}\times M_{{}_{{\text{H}}_2\text{O}}}$

Substitute $170\text{ mL}$ for $V_{{}_{{\text{H}}_2\text{O}}}$, $\text{1 g/mL}$ for $d_{{}_{{\text{H}}_2\text{O}}}$, $0.0510\text{ mol}$ for $n_{{}_{{\text{H}}_2\text{O}}}$, and $\text{18}\text{.016 g}$ for $M_{{}_{{\text{H}}_2\text{O}}}$ in the above equation

$\begin{array}{c}{m}_{{}_{{\text{H}}_2\text{O}}}=170\text{ mL}\times \text{1 g/mL}+0.0510\text{ mol}\times \text{18}\text{.016 g}\\ =170\text{ g}+0.9\text{ g}\\ \text{=170}\text{.9 g}\end{array}$

Substitute $\text{2}\text{.8662 kJ}$ for $q_{\text{surr}}$, $\text{4}\text{.184 J}/\text{g}{\text{.}}^{\text{o}}\text{C}$ for $s_{{\text{H}}_2\text{O}}$, $\text{170}\text{.9 g}$ for $m_{{}_{{\text{H}}_2\text{O}}}$, and ${1.67}^{\circ }\text{C}$ for $\Delta T$ in equation (1)

$\begin{array}{c}\text{2}\text{.8662 kJ}=\text{4}\text{.184 J}/\text{g}{\text{.}}^{\text{o}}\text{C}\times \text{170}\text{.9 g}\times {1.67}^{\circ }\text{C}+C_{\text{cal}}\times {1.67}^{\circ }\text{C}\\ \text{2}\text{.8662}\times {\text{10}}^3\text{ J=1194}\text{.12 J}+C_{\text{cal}}\times {1.67}^{\circ }\text{C}\\ \text{2866}\text{.2 J}-\text{1194}\text{.12 J=}{C}_{\text{cal}}\times {1.67}^{\circ }\text{C}\\ 1672.08\text{ J=}{C}_{\text{cal}}\times {1.67}^{\circ }\text{C}\end{array}$

So, the heat capacity of the calorimeter is given as

$\begin{array}{c}{C}_{\text{cal}}\text{=}\frac{1672.08\text{ J}}{{1.67}^{\circ }\text{C}}\\ =1001.2\text{ J}{/}^{\circ }\text{C}\\ \text{=1}\text{.00}\times {\text{10}}^3\text{ kJ}{/}^{\circ }\text{C}\end{array}$

Conclusion

The heat capacity of the calorimeter is calculated to be $\text{1}\text{.00}\times {\text{10}}^3\text{ kJ}{/}^{\circ }\text{C}$.