Which gives off more heat, a potato heated to 375°F or a 20-lb turkey heated to the same temperature? Explain your answer.

the heat emitted by potatoes is much less than the heat emitted by turkey

Explanation:

The heat transfer is given by

        Q = m ΔT

where m is the mass of the body, c_e is the specific heat and ΔT is the change in temperature of the body

in this case they ask us to appear the heat emitted by the potatoes and the turkey.

From the tables found the specious colors of both

potato     c_e = 0.82 Kcal / kg ºC

turkey      c_e = 0.72 Kcal / kg ºC

Since the two bodies have the same initial temperature and we assume that they reach the same final (ambient) temperature, the ratio of the heat emitted is

          Q₁ / Q₂ =  

where we use the rise 1 for the potatoes and the su index 2 for the turkey

the mass of the potatoes can be estimated at approximately m = 1 ka

we substitute

          Q₁ / Q₂ =  

           Q₁ /Q₂ = 0.0569

           Q₁ = 0.0569 Q₂

therefore the heat emitted by potatoes is much less than the heat emitted by turkey

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).

pretty sure it's false