Chapter 10, Problem 21A

Interpretation Introduction

Interpretation : The $\text{ΔE}$ for the system is to be calculated.

Concept Introduction : The change in the energy is calculated by the formula as shown below.

  $\Delta \text{E}=\text{q}+\text{w}$

Answer to Problem 21A

The $\text{ΔE}$ for the system is $-21\text{kJ}$ .

Explanation of Solution

The given value of heat is $\text{125}\text{kJ}$ . It is given that the system releases heat, which means the sign of heat is negative. Therefore,

  $\text{q}=-125\text{kJ}$

The given value of work done is $\text{104}\text{kJ}$ . It is given that the work is done on the system, which means the sign of work is positive. Therefore,

  $\text{w}=+104\text{kJ}$

Substitute the given value of heat and work done in the above formula.

  $\begin{array}{c}\Delta \text{E}=\text{q}+\text{w}\\ \text{=}-125\text{kJ}+104\text{kJ}\\ \text{=}-\text{21}\text{kJ}\end{array}$

Conclusion

The $\text{ΔE}$ for the system is $-21\text{kJ}$ .