Chapter 6, Problem 6.1P

Interpretation Introduction

Interpretation:

The internal energy change for the given system is to be calculated.

Concept Introduction: The internal energy change for the given system is calculated by using the formula,
   $\Delta E=q+w$

To determine: The internal energy change for the given system.

Answer to Problem 6.1P

Solution:

The internal energy change for the given system is $71\text{J}$ .

Explanation of Solution

Given Information:

The heat absorbed by the system is $559\text{J}$ .
The work done by the system is $488\text{J}$ .

The work done by the system has the negative sign and the heat absorbed by the system has the positive sign. Therefore, the heat change of the system is $+559\text{J}$ and the work done of the system is $-488\text{J}$ .

The internal energy change for the given system is calculated by using the formula,
   $\Delta E=q+w$

Where,

• $\Delta E$ is the internal energy change of the system.
• $q$ is the heat change of the system.
• $w$ is the work done of the system.

Substitute the values of $q$ and $w$ in the above equation.
   $\begin{array}{c}\Delta E=\left(+559\text{J}\right)+\left(-488\text{J}\right)\\ =71\text{J}\end{array}$

Conclusion

The internal energy change for the given system is $71\text{J}$ .