Chapter 6, Problem 6.53P

Interpretation Introduction

Interpretation:

The heat released $\left(q_{\text{v}}\right)$ per gram of hydrocarbon when $1.520\text{ g}$ of a hydrocarbon is burned in the bomb calorimeter that has a heat capacity of $11.09\text{ kJ/K}$ is to be determined.

Concept introduction:

A calorimeter is a device that is employed to measure the heat absorbed or released by the system. A calorimeter is of two types:

1. Coffee cup calorimeter

2. Bomb calorimeter

A bomb calorimeter is employed to determine the heat absorbed or released by the substance placed in the calorimeter at constant volume. A bomb calorimeter is generally used in case of combustion reactions. The heat lost by the reaction is gained by the calorimeter.

$-q_{\text{rxn}}=q_{\text{calorimeter}}$

Specific heat capacity $\left(c\right)$ of a substance is the amount of heat needed to raise the temperature of $1\text{g}$ of a substance by $1\text{K}$. The formula to calculate heat required is as follows:

$q=\left(\text{mass}\right)\left(c\right)\left(\Delta T\right)$ . (1)

Here,

$\Delta T$ is the temperature difference.

$q$ is the heat released or absorbed.

$c$ is the specific heat capacity of the substance.

The formula to calculate heat required in case of calorimeter is as follows:

$q_{\text{calorimeter}}=\left(C_{\text{calorimeter}}\right)\left(\Delta T_{\text{calorimeter}}\right)$ (2)

Here,

$\Delta T_{\text{calorimeter}}$ is the temperature difference.

$q_{\text{calorimeter}}$ is the heat released or absorbed by calorimeter.

$C_{\text{calorimeter}}$ is the specific heat capacity of the calorimeter.

At the constant volume, the value of heat $\left(q_{\text{v}}\right)$ is equal to internal energy $\left(\Delta E\right)$.

The formula to calculate $\Delta T$ of the system is as follows:

$\Delta T=T_{\text{Final}}-T_{\text{Initial}}$ (3)