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slava [35]
3 years ago
11

A metal entrance door swings freely in the winter, but when

Chemistry
1 answer:
ivann1987 [24]3 years ago
3 0

Answer:

This phenomenon occurs because the door, being metal and leading to changes in temperature, undergoes proportional and morphological changes, metals face expansion and expansion in the presence of heat, called thermal expansion.

On the other hand, against the cold, thermal contraction is suffered, that is why its volume decreases, and it contracts.

Explanation:

The expansion phenomenon of the door is not linear, since it increases its volume in width and height, therefore simultaneously on the entire surface.

When an area or surface expands, it does so by increasing its dimensions in the same proportion. For example, a metal sheet increases its length and width, which means an increase in area. Area dilation differs from linear dilation in that it involves an increase in area.

The area expansion coefficient is the increase in area that a body of a certain substance experiences, with an area equal to unity, as its temperature rises one degree centigrade. This coefficient is represented by the Greek letter gamma.

Regarding shrinkage, a clear example of this is when a metal foundry or a weld shrinks, sometimes it is difficult to understand with examples like these (doors) because it is little noticeable by our eyes and the dimensional changes for our perspective. it is infima.

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To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.

In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.

Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.

35.0° C - 25.0° C = 10.0° C

We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.

q = 125 g × 4.184 J/g °C × 10.0° C

q = 523 J/°C × 10.0° C

q = 5230 J

Therefore, it will take 5230 joules (J) to raise the temperature of the water.

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