Chapter 10, Problem 19A

(a)

Interpretation Introduction

Interpretation: If q is $75\text{kJ}$ and w is $-24\text{kJ}$ then $\Delta E$ 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 19A

The $\Delta E$ is $51\text{kJ}$ .

Explanation of Solution

According to question,

  $\begin{array}{l}\Delta E=?\\ q=75\text{kJ}\\ w=-24\text{kJ}\end{array}$

Substitute all values in internal energy expression as follows:

  $\begin{array}{c}\Delta E=q+w\\ =75\text{kJ}-24\text{kJ}\\ =51\text{kJ}\end{array}$

Hence, $\Delta E$ is $51\text{kJ}$ .

(b)

Interpretation Introduction

Interpretation: If q is $-54\text{kJ}$ and w is $-32\text{kJ}$ then $\Delta E$ 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 19A

The $\Delta E$ is $-86\text{kJ}$ .

Explanation of Solution

According to question,

  $\begin{array}{l}q=-54\text{kJ}\\ w=-32\text{kJ}\\ \Delta E=?\end{array}$

Substitute all values in internal energy expression as follows:

  $\begin{array}{c}\Delta E=q+w\\ =-54\text{kJ}-32\text{kJ}\\ =-86\text{kJ}\end{array}$

Hence, $\Delta E$ is $-86\text{kJ}$ .

(c)

Interpretation Introduction

Interpretation: If q is $-41\text{kJ}$ and w is $+118\text{kJ}$ then $\Delta E$ 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 19A

The $\Delta E$ is $77\text{kJ}$ .

Explanation of Solution

According to question,

  $\begin{array}{l}q=-41\text{kJ}\\ w=+118\text{kJ}\\ \Delta E=?\end{array}$

Substitute all values in internal energy expression as follows:

  $\begin{array}{c}\Delta E=q+w\\ =-41\text{kJ+118}\text{kJ}\\ =77\text{kJ}\end{array}$

Hence, $\Delta E$ is $77\text{kJ}$ .