Nuclear reactions in a reactor produce a lot of thermal

Chemistry

2 answers:

Anna11 [10]2 years ago

6 0

Answer:

Explanation:

Leokris [45]2 years ago

4 0

Answer:

Heat flows from the reactor to the water

Explanation:

The thermal energy mentioned in the description is another way to say heat. The energy that is produced by the nuclear reactions leaves the reactor and enters the water, warming it.

The passage does <em>not </em>say that heat flows in the form of electricity, but rather that the turbines turned by the steam produce electricity.

The passage does <em>not </em>say that the steam produces the heat, but rather that the boiling of the water (caused by the heat) produces steam.

You might be interested in

How were atomic models developed when no one had seen an atom?

AnnyKZ [126]

Atomic models were developed through indirect observation even though no one had seen an atom. There were many experiments of which I know nothing about, but in the end the scientists managed to come up with a formula and various models to describe atoms.

4 0

2 years ago

Explain how you determine the freezing point of a solution that does not have a well-defined transition in the cooling curve.

Sophie [7]

This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.

The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.

However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.

As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

$y=-3.5 x + 25\\\\y=-0.52 x + 2$

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

$-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72$

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

$y=-3.5 (7.72) + 25\\\\y = 1.84$

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.

Learn more: