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Sindrei [870]
3 years ago
9

A 115-g sample of steam at 100 °C is emitted from a volcano. It condenses, cools, and falls as snow at 0.0 °C. How many kilojoul

es were released?
Can someone please tell me how to do it step by step?
Chemistry
1 answer:
GarryVolchara [31]3 years ago
5 0

Answer:

∑Q(TTL) = 347 Kj. = 82.8 Kcal. (3 sig. figs.)

Explanation:

Draw a typical heating curve for water (see figure below) and label each section with data needed to calculate the amount of heat flow in the specified section.  

Super-imposing given data onto a trace of a typical heat flow chart for water finds there are three segments of the heating curve that can utilize the given data to calculate heat flow for each segment. With all three heat flow quantities, the total heat quantity associated with this problem is marked in orange. Calculate the heat quantity associated with the designated segments and add to obtain total heat flow for the transitions listed.  

Point E to D => Q₄ = m∙ΔHᵥ = (115g)(540 cal/g) = 62,100 cals.

Point D to C => Q₃ = m∙c∙ΔT = (115g)(1 cal/g∙˚C)(100˚C) = 11,500 cals.

Point C to B => Q₂ = m∙ΔHₓ = (115g)(80cal/g) = 9,200 cals.

Total Heat Flow (∑Qₙ) = Q₄ + Q₃ + Q₂ = 62,100 cals.  +  11,500 cals.  +  9,200 cals. = 82,800 cals.

= (82,800 cals.)(4.184 joules/cal.) = 346,435 joules = 346.435 Kj ~ 347 Kj (3 sig. figs.) _____________________________________________________

m = mass = 115g

c = specific heat of liquid water = 1 calorie/gram·°C = 4.184 j/g·°C

ΔT = temperature change in degrees Celsius

ΔHᵥ = heat of vaporization = 540 cals./gram

ΔHₓ = heat of crystallization = 80 cal./gram

Qₙ = heat flow quantity per specific segment (calories or joules)

∑Qₙ = total heat flow

Heating curve for water:

Note in the diagram that only two formulas are used.

Q = m·c·ΔT => heating or cooling the pure condensed state. The segments demonstrate temperature change.

Q = m·ΔHₙ => heat flow during phase change. Note, in these segments of the heating curve two phases are in contact. That is solid/liquid or liquid/gas phase substances. Also note, in these segments, when two phases are in contact no temperature change occurs. Examples, ice water remains at a constant temperature until all ice is melted or all of the liquid water is frozen depending upon the direction of heat flow. The same is true for boiling water in that when two phases are in contact (liquid/gas), temperature remains constant. The portions of the heating curve designating phase transitions are horizontal and are defined by the equation Q = m·ΔH,  while the curve segments that are only one phase demonstrate temperature change and are defined by the equation containing temperature change, Q = m·c·ΔT.  

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Answer:

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Explanation:

It seems the question is incomplete. But an internet search shows me these values for the question:

" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."

Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>

First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:

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The<u> mole fraction of thiophene</u> is:

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Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene

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