Chapter 8, Problem 16QAP

Ethyl alcohol, C₂H₅OH, is the intoxicating agent in liquor. Burning 1.00 g of ethyl alcohol in an excess of oxygen at 23.28°C and constant volume releases 29.52 kJ of heat. When 7.40 g of ethyl alcohol is burned in oxygen under the same conditions in a bomb calorimeter, the temperature of the bomb and water rises from 23.28°C to 48.04°C. The bomb holds 0.750 kg of water.

(a) What is q for the combustion of the ethyl alcohol in the bomb calorimeter?

(b) What is qH₂O?

(c) What is q_cal?

(d) What is the heat capacity of the calorimeter?

Interpretation Introduction

(a)

Interpretation:

To determine q for the combustion of ethyl alcohol in bomb calorimeter.

Concept introduction:

In a bomb calorimeter, heat given off by the reaction is absorbed by the calorimeter (including metal bomb and water). Therefore,

${\text{q}}_{\text{reaction}}\text{=-}\left({\text{q}}_{\text{calorimeter}}\text{+}{\text{q}}_{\text{water}}\right)$

Heat absorbed by bomb calorimeter can be calculated by using below formula:

$\begin{array}{l}{\text{q}}_{\text{calorimeter}}\text{=}\text{CΔT}\\ \text{where,}\\ \text{C = specific heat of calorimeter}\\ \text{ΔT}\text{=}\text{change in temperature}\end{array}$

Heat absorbed by water can be calculated by using below formula:

$\begin{array}{l}{\text{ q}}_{\text{water}}\text{=}\text{mCΔT}\\ \text{where,}\\ \text{m=}\text{mass}\text{of}\text{water}\\ \text{C=}\text{specific heat of water}\\ \text{ΔT= change in temperature}\\ \end{array}$

Answer to Problem 16QAP

q for combustion of 7.40 g of ethyl alcohol is -218.44 kJ.

Explanation of Solution

According to question:

Combustion of 1 g of ethyl alcohol releases 29.52 kJ energy.

Therefore,

$\begin{array}{l}\text{Combustion of 7}\text{.40 g of ethyl alcohol will}\text{releases }\left(\text{29}\text{.52×7}\text{.40}\right)\text{ kJ energy}\text{.}\\ \text{=218}\text{.44}\text{kJ}\end{array}$

q for combustion of 7.40 g of ethyl alcohol is -218.44 kJ.

Here negative sign indicates that energy is released.

Interpretation Introduction

(b)

Interpretation:

To determine q_H2O in the given experiment.

Concept introduction:

In a bomb calorimeter, heat given off by the reaction is absorbed by the calorimeter (including metal bomb and water). Therefore,

${\text{q}}_{\text{reaction}}\text{=-}\left({\text{q}}_{\text{calorimeter}}\text{+}{\text{q}}_{\text{water}}\right)$

Heat absorbed by bomb calorimeter can be calculated by using below formula:

$\begin{array}{l}{\text{q}}_{\text{calorimeter}}\text{=}\text{CΔT}\\ \text{where,}\\ \text{C = specific heat of calorimeter}\\ \text{ΔT}\text{=}\text{change in temperature}\end{array}$

Heat absorbed by water can be calculated by using below formula:

$\begin{array}{l}{\text{ q}}_{\text{water}}\text{=}\text{mCΔT}\\ \text{where,}\\ \text{m=}\text{mass}\text{of}\text{water}\\ \text{C=}\text{specific heat of water}\\ \text{ΔT= change in temperature}\\ \end{array}$

Answer to Problem 16QAP

q_H2O is 77.62 kJ.

Explanation of Solution

According to question:

Mass of water = 0.750 kg = 750 g

Initial temperature = 23.28 °C

Final temperature = 48.04 °C

Change in temperature = 48.04 - 23.28 = 24.76 °C

Specific heat of water = 4.18 J/g°C

As we know:

$\begin{array}{l}{\text{ q}}_{\text{water}}\text{=}\text{mCΔT}\\ \text{where,}\\ \text{m=}\text{mass}\text{of}\text{water}\\ \text{C=}\text{specific heat of water}\\ \text{ΔT= change in temperature}\\ \end{array}$

Plugging values, we get:

$\begin{array}{l}{\text{ q}}_{\text{water}}\text{=}\text{mCΔT}\\ {\text{q}}_{\text{water}}\text{=}\text{750×4}\text{.18×24}\text{.76}\\ {\text{q}}_{\text{water}}\text{=}\text{77622}\text{.6}\text{J}\\ {\text{q}}_{\text{water}}\text{=}\text{77}\text{.62}\text{kJ}\end{array}$

Thus,q_H2O is 77.62 kJ.

Interpretation Introduction

(c)

Interpretation:

To determine q_calorimeter in the given experiment.

Concept introduction:

In a bomb calorimeter, heat given off by the reaction is absorbed by the calorimeter (including metal bomb and water). Therefore,

${\text{q}}_{\text{reaction}}\text{=-}\left({\text{q}}_{\text{calorimeter}}\text{+}{\text{q}}_{\text{water}}\right)$

Answer to Problem 16QAP

q_calorimeter is 140.82 kJ.

Explanation of Solution

Here we have:

q_H2O = 77.62 kJ.

q_reaction = -218.44 kJ. (enthalpy of reaction)

As we know:

${\text{q}}_{\text{reaction}}\text{=-}\left({\text{q}}_{\text{calorimeter}}\text{+}{\text{q}}_{\text{water}}\right)$

Plugging values, we get:

$\begin{array}{l}{\text{q}}_{\text{reaction}}\text{=-}\left({\text{q}}_{\text{calorimeter}}\text{+}{\text{q}}_{\text{water}}\right)\\ \Rightarrow \text{-218}\text{.44}\text{=}\text{-}\left({\text{q}}_{\text{calorimeter}}\text{+77}\text{.62}\right)\\ \Rightarrow \text{-218}\text{.44}\text{=}{\text{-q}}_{\text{calorimeter}}\text{-77}\text{.62}\\ \Rightarrow {\text{q}}_{\text{calorimeter}}\text{=}\text{-77}\text{.62+218}\text{.44}\\ \Rightarrow {\text{q}}_{\text{calorimeter}}\text{=}\text{140}\text{.82kJ}\end{array}$

Thus, q_calorimeter is 140.82 kJ.

Interpretation Introduction

(d)

Interpretation:

To determine the heat capacity of calorimeter in the given experiment.

Concept introduction:

Heat absorbed by bomb calorimeter can be calculated by using below formula:

$\begin{array}{l}{\text{q}}_{\text{calorimeter}}\text{=}\text{CΔT}\\ \text{where,}\\ \text{C = specific heat of calorimeter}\\ \text{ΔT}\text{=}\text{change in temperature}\end{array}$

Answer to Problem 16QAP

Heat capacity of bomb calorimeter is 5.68 kJ/°C

Explanation of Solution

Here we have:

q_calorimeter = 140.82 kJ.

Change in temperature = 24.76°C

As we know:

$\begin{array}{l}{\text{q}}_{\text{calorimeter}}\text{=}\text{CΔT}\\ \text{where,}\\ \text{C = specific heat of calorimeter}\\ \text{ΔT}\text{=}\text{change in temperature}\end{array}$

Plugging values, we get:

$\begin{array}{l}{\text{q}}_{\text{calorimeter}}\text{=}\text{CΔT}\\ \Rightarrow \text{140}\text{.82}\text{=}\text{C×24}\text{.76}\\ \Rightarrow \text{C}\text{=}\text{5}\text{.68}\text{kJ/°C}\end{array}$

Thus, heat capacity of bomb calorimeter is 5.68 kJ/°C