Chapter 10, Problem 20A

Interpretation Introduction

Interpretation: A gas absorbs 45 kJ of heat and 29 kJ of work the value of $\Delta E$ needs to be explained.

Concept Introduction: The internal energy E of a system is altered by a work flow, heat or both. Mathematically it is represented as:

  $\Delta E=q+w$

Here, $q$ and $w$ represents heat and work respectively and the symbol $\Delta$ indicates change in the function that follows.

Answer to Problem 20A

The $\Delta E$ value is $16\text{kJ}$ .

Explanation of Solution

Calculate $\Delta E$ as follows:

As heat is absorbed so q value is positive. Work is done by the system so, w value is negative. Therefore, given values are as follows:

  $\begin{array}{c}q=45\text{kJ}\\ w=-29\text{kJ}\\ \Delta E=?\end{array}$

Put all values in below equation as follows:

  $\begin{array}{c}\Delta E=q+w\\ =45\text{kJ}-29\text{kJ}\\ =16\text{kJ}\end{array}$

Therefore, $\Delta E$ is $16\text{kJ}$ .