Calculate the rate of heat generation in kW due to the burning of the fuel when you drive a car rated at 34 MPG (miles per gallon) at a speed of 70 MPH (miles per hour), given that the heating value of gasoline, that is the heat generated when a unit mass of gasoline is burnt in air is 44 x 103 kJ/kg.

Explanation:

Let us calculate gallon used in one hour .

It travels 70 miles in one hour

in 70 miles it uses 70 / 34 gallons of fuel

70 / 34 gallons = 70 / 34 x 3.7854 kg

= 7.8 kg

heat generated = 7.8 x 44 x 10⁶ J

= 343.2 x 10⁶ J

This is heat generated in one hour

heat generated in one second = 343.2 x 10⁶ / 60 x 60 J/s

= 95.33 x 10³ J /s

= 95.33 kW.

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).