In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the system if the initial energy was 2000 j? 1. write the equation 2.list out your known variables 3.plug the numbers into the equations 4.solve 5.write your solution statement that includes initial energy and final

2500 J

Explanation:

The first law of thermodyniamics states that:

where

is the change in internal energy

Q is the heat absorbed by the system

W is the work done by the system

in this problem, Q=1000 J and W=500 J, so we have

Since the initial internal energy is , we the final energy is given by:

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).