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siniylev [52]
3 years ago
9

Devin investigated whether salt reduces the boiling point of water. She poured 1 gallon of 18 degrees Celsius (°C) water into a

pot and added 2 cups of salt to it. Devin heated it over a fixed flame and measured the temperature when the water reached a rolling boil. She repeated the experiment with 1 gallon of water and 4 cups of salt under identical conditions. Where did Devin make a mistake in her experimental design?
She did not boil a pot of water without salt.


She waited until the water reached a rolling boil.


She used different amounts of salt in the two pots.


She did not heat the pots of water at the same time.
Physics
2 answers:
ankoles [38]3 years ago
6 0
Heheh I’m dying help call 911
daser333 [38]3 years ago
4 0

Answer:C

Explanation:

She used 2 different amounts of salts so that was her mistake she should of used 1

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What is the meaning of critical angle in physics​
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A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo
LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But  ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

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4 years ago
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3 years ago
Read 2 more answers
Can someone please help?
OLga [1]

Answer:

1: E

2: A

3: C

4: B

5: D

Explanation:

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