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bearhunter [10]
3 years ago
14

A heat engine is designed to do work. This is possible only if certain relationships between the heats and temperatures at the i

nput and output hold true. Which of the following sets of statements must apply for the heat engine to do work?
A) Qh < Qc and Th < Tc
B) Qh > Qc and Th < Tc
C) Qh < Qc and Th > Tc
D) Qh > Qc and Th > Tc

Part B
Find the work W done by the "ideal" heat engine.
Express W in terms of Qh and Qc
W=___________

Part C
The thermal efficiency e of a heat engine is defined as follows: e = W/Qh
Express the efficiency in terms of Qh and Qc.
e=____________
Physics
1 answer:
Tcecarenko [31]3 years ago
4 0

Answer:

Part A: D

Part B: W = Qh - Qc

Part C: e = 1 - Qc/Qh

Explanation:

The heat engine is the engine that transforms heat (Q) in work (W), and by the second law of the thermodynamics, its efficiency can not be 100%, it means that some heat must be dissipated.

Part A:

The engine works with two sources of heat, one hot (Qh) at a hot temperature (Th) and another cold (Qc) at a cold temperature (Tc). It is necessary so, the hot source will give energy to the fluid of the engine, and the cold source will be the source where these heat will dissipate and the fluid will return to its original temperature. So,

Qh > Qc, and Th > Tc

Part B:

The ideal heat engine is the one that can use the most amount of heat to transform it at work. It is characterized by Qh/Qc = Th/Tc.

The work is the useful energy, so it is the total heat (Qh) less the heat dissipated (Qc):

W = Qh - Qc

Part C:

The effiency is the useful energy divided by the total energy. Because W = Qh - Qc:

e = W/Qh

e = (Qh - Qc)/Qh

e = Qh/Qh - Qc/Qh

e = 1 - Qc/Qh

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A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i
Effectus [21]

Answer:

s =  1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore  

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}  

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}  

• s = \sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}

s = \sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}

s =  1.7 m

7 0
3 years ago
The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
In a binary star system, an unseen component is found to have 8 solar masses. It would be visible if the system were a normal st
maksim [4K]

Answer:

Black Hole

Explanation:

A black hole is a very dense and massive stellar object, which has a field of gravity so large that not even light can escape it.

Since it does not emit light, <u>we cannot see them directly</u>, hence the name of black hole.

So in this case,<u> if the object has a mass of 8 solar masses that is enough to form a black hole</u>, and <u>also cannot be seen</u>, all of this indicates that the object we are talking about is a black hole.

It should be mentioned that although these objects do not emit light, because it cannot escape due to the immense force of gravity, black holes can be detected by a type of radiation emitted on their event horizon due to quantum effects called Hawking radiation .

5 0
3 years ago
Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o
VARVARA [1.3K]

The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so,  we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

K=Celsius+273.15

So, converting we have:

T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K

Therefore, substituting the given information and calculating, we have:

HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )

HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0
3 years ago
Suppose the three resistors in the picture represents a 240 W
Airida [17]

A fuse is an electrical safety device which should not blow, which should overheat and melts if current is too high. Its placed in the live wire before the switch. This prevents overheating and catching fire. A fuse have a specific current value for example - 3000 amps. So when choosing a suitable fuse you must use the above minimum value but less than maximum value. For example in a circuit there is 1000W flowing, you should choose more than 1000 amps fuse not less or else, it will melt.

3 0
3 years ago
Read 2 more answers
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