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pochemuha
3 years ago
14

A 1.0 liter pot is filled with water at sea level and is brought to a boil. the same pot is filled with water at 5,000 feet abov

e sea level and is brought to a boil. assuming the same heat is being applied to the pots and that they are filled with the same amount of water, which pot will boil faster and why?
Physics
1 answer:
kykrilka [37]3 years ago
8 0
The second pot will boil faster

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Answer:

the answer is c

Explanation:

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You may make the following measurements of an object 42kg and 22m3. What would the objects density be?
ki77a [65]
Density=mass/volume
to find the density
mass=42kg
volume=22m3
so density=42/22
density=1.9Kgm3
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Can an object contain heat? Why or why not?
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Every object has thermal energy (better word than heat, since we associate that with high temperatures). This is actually the molecules vibrating, moving a lot. More thermal energy means more vibrating, and thus also expanding in volume.
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The way the brain perceives the intensity of a sound is
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Read 2 more answers
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
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