Chapter 10, Problem 21A

Interpretation Introduction

Interpretation: A system releases 213 kJ of heat and calculated $\Delta E$ is $-\text{45}\text{kJ}$ , then work done on the system 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 21A

The work done (w) on the system is $\text{168}\text{kJ}$ .

Explanation of Solution

Calculate w as follows:

As heat is releases so, q value is negative. Therefore, given values are as follows:

  $\begin{array}{c}q=-213\text{kJ}\\ \Delta E=-45\text{kJ}\\ w=?\end{array}$

Substitute all values in below equation.

  $\begin{array}{c}\Delta E=q+w\\ -45\text{kJ}=-213\text{kJ+}w\\ w=-45\text{kJ+}213\text{kJ}\\ \text{=168}\text{kJ}\end{array}$

Therefore, w is $\text{168}\text{kJ}$ .