Drawing heating cooling curve, Phase change in boiling pasta water

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Dhamnekar Winod
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Drawing heating cooling curve, Phase change in boiling pasta water

Post by Dhamnekar Winod »

1)Draw a heating cooling curve for a substance that freezes at -57° C and boils at 125.6° Label each phase of the curve with its appropriate name. Also

indicate whether each change is endothermic or exothermic.

2)Your friend asks you what temperature would you like to use to boil your pasta water. Write out how you would explain the idea of phase change to your friend, and teach them about temperature during a phase change.

How to answer both these questions. What are the answers to both these questions? I am working on it.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Dhamnekar Winod
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Re: Drawing heating cooling curve, Phase change in boiling pasta water

Post by Dhamnekar Winod »

Answer to first question given by Chemistry professor as follows:-
Supposing the substance in question is n-octane, which has freezing and boiling points very similar to those given in this question.

The given freezing and boiling points determine the vertical position of the two horizontal portions of the graph. The length of these horizontal lines is determined by the heats of fusion and vaporization, which are 20.7 and 44.0 kJ/mol respectively.

The slope of the tilted lines connecting the horizontal lines are determined by the heat capacities of the solid, liquid, and gas phases which are 239, 255, and 233 J/mol·K, respectively.

Supposing the graph is going to show the values for a sample of exactly 1 mole of octane at constant pressure. Calculating the inflection points on the graph:

Starting arbitrarily at -150°C: (0, -150)

Beginning of the melting:

(-57 - (-150))K x (239 J/K) = 22227 J = 22.2 kJ : (22.2, -57)left to right

End of the melting:

22.2 kJ + 20.7 kJ = 42.9 kJ : (42.9, - 57)

Beginning of the vaporization:

(255 J/K x (125.6 - (-57))K) = 46563 J = 46.6 kJ added

42.9 kJ + 46.6 kJ = 89.5 kJ : (89.5, 125.6)

End of the vaporization:

89.5 kJ + 44.0 kJ = 133.5 kJ : (133.5, 125.6)

Ending arbitrarily at 200°C:

(233 J/K) x (200 - 125.6)K = 17335.2 J = 17.3 kJ added

133.5 kJ + 17.3 kJ = 150.8 kJ : (150.8, 200)

[Notice that all heating curves have this general shape, two plateaus connected to three lines that have positive slopes. The only difference among substances is where the inflection points are located.]

Any process which moves on the graph from left to right is endothermic (heat absorbed).

Any process which moves on the graph from right to left is exothermic (heat released).

Image

Source(s):https://webbook.nist.gov/cgi/cbook.cgi? ... 659&Mask=4


Answer to second question is as follows:-

You can't get the temperature of boiling water much over 212F or 100C. The water turns to vapor at that temperature. The temperature will remain at 212F or 100C until all the water has boiled away into the surrounding atmosphere.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Drawing heating cooling curve, Phase change in boiling pasta water

Post by ChenBeier »

You post how to draw a curve of melting and boiling a substance. Also told the professors solution.

But what is now unclear for you?
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