Chapter 10, Problem 1SAQ

A chemical system produces 155 kJ of heat and does 22 kJ of work. What is ΔE for the surroundings?

1. -177 kJ
2. 177 kJ
3. -133 kJ
4. 133 kJ

Interpretation Introduction

Interpretation:

To find out $\text{ΔE}$ for the surroundings of a chemical system that produces $155\text{kJ}$ of heat and does $22\text{kJ}$ of work.

Concept Introduction: The relationship between heat, $q$ , work, $w$ , and $\Delta E_{\text{sys}}$ is given as:

  $\Delta E_{\text{sys}}=q+w$

Answer to Problem 1SAQ

Correct answer:

Therefore, option (b) is correct and options (a), (c) and (d), are incorrect.

Explanation of Solution

Reasons for the correct statement: The heat is produced by the chemical system and released to the surroundings so, $\text{q}\text{=}\text{-}\text{155}\text{kJ}$

The work is done by the chemical system on the surroundings so, $w\text{=}\text{-}\text{22}\text{kJ}$

Substituting the values:

  $\begin{array}{l}\Delta E_{\text{sys}}=q+w\mathrm{.........}\left(1\right)\\ \Delta E_{\text{sys}}=-155\text{kJ}+\left(-22\text{kJ}\right)\\ =-177kJ\\ \Delta E_{\text{sys}}=-\Delta E_{\text{surr}}(\text{or})\\ \Delta E_{\text{surr}}=-\Delta E_{\text{sys}}\\ =-\left(-177\text{kJ}\right)\\ =177\text{kJ}\end{array}$

Hence, option (b) is correct.