Question

A heat engine satisfies $W_{out}=Q_{out}$ Why is there no $\mathrm{\Delta }{E}_{\text{th}}$ term in this relationship?

Short answer

Since it is zero.

Step-by-step solution

Step :1 Introduction 

The first law of thermodynamics is a thermodynamic adaptation of the concept of conservation of energy, differentiating three types of energy transfer: heat, thermodynamic work, and energy associated with matter transfer, and linking these to an internal energy function of a body.

The law of conservation of energy asserts that the total energy of any isolated system (in which energy and matter cannot be transferred beyond the system border) is constant; energy can be changed from one form to another, but it cannot be created or destroyed.

Step :2 Explanation 

First law of thermodynamics

$\mathrm{\Delta }{E}_{th}=W+Q$

where $w$is the work done on the gas. The work done by the gas is exactly equal to the negative of the work done on the gas. That is, we can write the first law as

$\mathrm{\Delta }{E}_{\text{th}}=W_{\text{out}}+Q_{\text{net}}$

Step :3  Thermic cycle

The author has denoted $Q_{net}$, This is the total amount of heat transferred to the gas. Heat is delivered to and withdrawn from the gas in a thermic cycle. Please keep in mind that, because heat engines function in cycles, we consider one cycle to be spinning in the same direction. $(p,v)$The temperature of the gas, and hence its internal energy, is clearly the same after one cycle as it was before the cycle..