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3 years ago

A heat engine operating between energy reservoirs at 20?c and 600?c has 30% of the maximum possible efficiency.

1 answer:

8 0

The maximum possible efficiency, i.e the efficiency of a Carnot engine , is give by the ratio of the absolute temperatures of hot and cold reservoir.
η_max = 1 - (T_c/T_h)

For this engine:
η_max = 1 - [ (20 +273)K/(600 + 273)K ] = 0.66 = 66%

The actual efficiency of the engine is 30%, i.e.
η = 0.3 ∙ 0.664 = 0.20 = 20 %

On the other hand thermal efficiency is defined as the ratio of work done to the amount of heat absorbed from hot reservoir:
η = W/Q_h

So the heat required from hot reservoir is:
Q_h = W/η = 1000J / 0.20 = 5000J

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A baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg. What’s the baseballs velocity?

Mashcka [7]

Answer:

v=40 m/s south

Explanation:

Momentum

It's a physical magnitude that measures the product of the mass by the velocity of a particle. Its units in the International System is kg.m/s and the formula is

$p=m.v$

Where m is the mass and v the velocity of the particle. If we wanted to solve for v, we have

$\displaystyle v=\frac{p}{m}$

The baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg, thus

$\displaystyle v=\frac{6}{0.15}=40\ m/s$

The velocity is directed to the south

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4 years ago

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

What happenes to our body temperature on a cold winter day or a hot summer day

viva [34]

In cold winter day, the body temperature falls down from normal temperature of 98.6°F (37°C) to 95°F (35°C). In winter body losses heat faster than it generates heat. If the temperature fall further below 95°F (35°C), it is emergency condition known as Hypothermia. One has to consult doctor in this case.

In summer hot days, body evaporates water in the form of sweat, in order to remain itself cool. Rise of temperature up to 100°F is normal. It is recommended to hydrate body to maintain temperature in summer days.

3 0

3 years ago

During a hurricane in 2008, the Westin Hotel in downtownNew Orleans suffered damage. Suppose a piece of glass dropped near the t

tangare [24]

Answer:

77.8 m/s, downward

Explanation:

For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf

Average speed = (Vf + Vi)/2

Also, by definition, the average speed is the distance divided by the time:

Average speed = distance / time

Then:

(Vf + Vi)/2 = 300m/6.62s

Other kinematic equation for uniform acceleration is:

Vf = Vi + a×t

Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)

Replacing the known values we can set a system of two equations:

From (Vf + Vi)/2 = 300m/6.62s

(Vf + Vi) = 2 × 300m/6.62s

Vf + Vi = 90.634 equation 1

From Vf = Vi + a×t

Vf - Vi = 9.8 (6.62)

Vf - Vi = 64.876 equation 2

Adding the two equations:

2Vf = 155.510

Vf = 77.8 m/s downward (velocities must be reported with their directions)

8 0

3 years ago

A 12.0-g plastic ball is dropped from a height of 2.50 m. Just as it strikes the floor, it is moving at a speed of 3.20 m/s. How

nalin [4]

Answer:

0·233 J

Explanation:

Given

Mass of the ball = 0·012 kg

Initially the ball is at a height of 2·5 m

As initially the ball is dropped, it's initial velocity will be equal to 0

Therefore initially it has zero kinetic energy and has only potential energy

∴ Initially total mechanical energy of the ball = potential energy of the ball

Initial potential energy of the ball = m × g × h

where

m is the mass of the ball

g is the acceleration due to gravity

h is the height of the ball

∴ Potential energy = 0·012 × 9·8 × 2·5 = 0·294 J

Velocity of the ball after striking the floor = 3·2 m/s

After striking the floor, the total mechanical energy = kinetic energy just after striking the floor

Kinetic energy = 0·5 × m × v²

where m is the mass of the ball

v is the velocity of the ball

∴ Kinetic energy of the ball = 0·5 × 0·012 × 3·2² = 0·061 J

Mechanical energy that is lost = 0·294 - 0·061 = 0·233 J

∴ Mechanical energy that the ball lost during its fall = 0·233 J

6 0

3 years ago