How much heat in kJ must be transferred to 10 kg of air to increase the temperature from 10C to 230C if the pressure is maintained constant? a.2310 b.1980 c.1650 d.2200

Answers

Answer 1

The amount of heat that must be transferred to 10 kg of air to increase the temperature from 10C to 230C at constant pressure is 2200 kJ. The correct option is d.2200.


The amount of heat required to increase the temperature of a substance can be calculated using the specific heat capacity (c) of the substance, its mass (m), and the change in temperature (ΔT) using the formula Q = mcΔT.

For air at a constant pressure, the specific heat capacity is approximately 1.005 kJ/kg·K. Therefore, to calculate the amount of heat required to increase the temperature of 10 kg of air from 10C to 230C, we first need to calculate the change in temperature:

ΔT = (230C - 10C) = 220C

Then, we can use the formula to find Q:

Q = mcΔT = (10 kg) x (1.005 kJ/kg·K) x (220C) = 2200 kJ

Therefore, the amount of heat that must be transferred to 10 kg of air to increase the temperature from 10C to 230C at constant pressure is 2200 kJ. The correct option is d.2200.

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Related Questions

The need for extraction of raw metals for making steel has been reduced due to the?

Answers

Answer: Increase in minimills

Explanation:

What frequency bands are used primarily for short range distances select all the nearby?

Answers

The frequency bands that are used primarily for short-range distances include the following:

The upper portion of the VHF band (225 MHz to 300 MHz).The lower portion of the UHF band (300 MHz to 400 MHz).

What is GSM?

GSM is an abbreviation for global system for mobile communication and it can be defined as an open and digital cellular network technology that is designed and developed to be used for the transmission of both mobile voice and data services at the same time.

In Computer technology and networking, the global system for mobile communication (GSM) is typically designed and developed to operates at the following bands of frequencies:

850 MHz900 MHz1800 MHz1900 MHz

In this context, we can infer and logically deduce that the frequency bands that are used primarily for short-range distances such as PAN, include the following:

The upper portion of the VHF band (225 MHz to 300 MHz).The lower portion of the UHF band (300 MHz to 400 MHz).

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A +7.5% grade meets a horizontal grade on a section of a rural mountainous highway. If the length of the crest vertical curve formed in that section is 300 ft long, determine the safe operating speed on the highway.

Answers

Using the formula of comfort condition it is possible to determine the safe operating speed on the highway.

What is a deflection angle?

In physics, we define angular deviation as the angle resulting from the deflection of light when reflected from a prism. A prism has the ability to separate white light into several other colors, which are reflected by the object.

The comfort condition formula is given by: [tex]L=2(\frac{NV^{3} }{C} )^{\frac{1}{2} }[/tex]

Where, L is the length of the vertical curve, N is the grade, V is the safe speed operation and assuming the standart value for C (0,6 m/s3) it is possible to calculate:

[tex]C = 0,6 m/s^{3} = 1,968 ft/s^{3}\\ N= N1 - N2 = 7,5-0 = 7,5% = 0,075[/tex]

Changing the values in the formula:

[tex]L=2(\frac{NV^{3} }{C}){\frac{1}{2} } \\300 = 2(\frac{0,075.V^{3} }{1,968})^{\frac{1}{2} } \\V = 83,89 ft/s[/tex]

So, the safe operating speed on the highway is 83,89 ft/s.

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The high-pressure fuel pumps used in gasoline direct-injection (GDI) systems are powered by ________.

Answers

Answer:

Camshaft

Explanation:

In order to live and grow, bacteria need moisture, food, the right temperature, and ______? Fill in the blank

Answers

Answer:

acidity, time,oxygen

Explanation:

These are some conditions for the growth of bacteria

Consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the pressure drop will _____.

Answers

The answer is 6because I just took the test!

different lever designs can be engineered to alter the brake pedal effort required of the driver by using different levels of ?

Answers

Different lever designs can be engineered and developed to alter the brake pedal effort required of the driver by using different levels of mechanical advantage.

What is mechanical advantage?

Mechanical advantage can be defined as a ratio of the output force of a lever to the force acting on it (input force or effort), assuming no losses due to wear, flexibility, tear or friction.

This ultimately implies that, different lever designs can be suitably engineered and developed to alter the brake pedal effort (input force) that is required of the driver, especially by using different levels of mechanical advantage.

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With _____, only one criterion must evaluate true in order for a record to be selected and with _____, all criteria must be evaluate true in order for a record to be selected.

a. parameter criteria, double criteria
b. function criteria, IF criteria
c. simple criteria, complex criteria
d. OR criteria, AND criteria

Answers

Answer:

d

Explanation:

OR criteria, AND criteria

In an OR criteria, it doesn't need all the records to be true. Just one record is enough and all other criterion becomes true.

In an AND criteria, it's unlike the OR criteria and works in opposite. It needs every member of the record to be true to be able to adjudge the whole record as true.

And as such, we have

With OR criteria, only one criterion must evaluate true in order for a record to be selected and with AND criteria, all criteria must be evaluate true in order for a record to be selected.

True or false? if i were to hook up an ac voltage source to a resistor, the voltage drop across the resistor would be in phase with the current in the circuit.

Answers

Answer: True

Explanation:

how skateboards works?

Answers

Answer

The skateboarder applies pressure to the trucks and gives/releases pressure on the levers. Second, the wheels and the axles are also examples of simple machines. They help the skater ride, spin, grind, and do a bunch of other radical movements on a skateboard.:

Explanation:

An example of a power consuming device would be a(n) ____ while an example of a non-power consuming device would be a(n) ____.

Answers

An example of a power consuming device would be a(n) light bulb, a computer device while an example of a non-power consuming device would be a(n) switch or button.

What is power consumption?

The quantity of energy utilized per unit of time is known as power consumption. Power use is a significant factor in digital systems.

Power consumption is a limiting factor for the battery life of portable devices like laptop computers and cell phones.

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Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 399 MPa causes the specimen to plastically elongate to a length of 54.4 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 47.7 mm to a length of 57.8 mm.

Answers

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

[tex]\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}[/tex]

Solve the following for the real stress and pressure for the stable.[tex]\sigma_{r1}=K(\varepsilon_{r1})^{n}[/tex]

[tex]K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}[/tex]

Solve the following for the true state stress and stress2.

[tex]\sigma_{r2}=K(\varepsilon_{r2})^n[/tex]

     [tex]=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa[/tex]

Explain any three ways research can facilitate the work of building technicians

Answers

According to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.

What are building technicians?

They are experts in construction projects, supporting the design of plans, estimating costs, planning work methods, etc.

Research can facilitate the work of building technicians in the following ways:

Conception, design and planning of the project.Selection of materials for floor, wall and ceiling systems.Carry out the programming of works, budgets and analysis of unit prices.

Therefore, we can conclude that according to the research, research can facilitate the work of building technicians in the conception, materials to use and planning of the project.

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This is silence I couldent find the tab... 30 points plus marked brainliest if corrects!


The most recent evidence supporting the theory of plate tectonics would include
es )
A)
GPS monitoring of plate speeds and movements.
B)
the WWII discovery of paleomagnetic reversals.
Elimi
O
the 1963 mapping of the tectonic plate boundaries.
D
C-14 dating of marine fossils found in the Himalayas.

Answers

Yeah the answer is A

I have an AC waveform with the voltage peak value of 100 V. I'm trying to find the RMS value of the voltage​

Answers

Answer:

98 x 100=9,00

Explanation:

What are the top 4 solar inventions, how they are used, and how they are better than the original way of powering them

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Yes I will answer soon

what is the distance in term of wavelengh between successive minima in the standing wave ratio​

Answers

Answer:

hi there

Explanation:

It is possible to design a reactor where the scy conductor and the nitrogen/ammonia electrode operate at different temperatures. Which combination of temperatures is expected to give the best results?.

Answers

The combination of temperatures that is expected to give the best result is option A: SCY temperature higher than electrode temp.

Why the case above?

The answer is option A because the proton conductivity of SCY is known to be one that often goes up along with increasing temperature, and the favorability of reaction goes down or reduces with total temperature.

Hence, It is said to be good for one to to keep the SCY conductor at a higher temperature and as such, The combination of temperatures that is expected to give the best result is option A: SCY temperature higher than electrode temp.

See options below

. Which combination of temperatures is expected to give the best results?

A.

SCY temperature higher than electrode temperature

B.

SCY temperature lower than electrode temperature

C.

SCY temperature the same as electrode temperature

D.

The temperature of the components does not make a difference.

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Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligible medium resistance. The cake solids (dry basis) per volume of filtrate was 20 g/liter. It is desired to operate a larger rotary vacuum filter (diameter 8 m and length 12 m) at a vacuum pressure of 80 kPA with a cake formation time of 20 s and a cycle time of 60 s. Determine the filtration rate in volumes/hr expected for the rotary vacuum filter.

Answers

Answer:

5.118 m^3/hr

Explanation:

Given data:

viscosity of cell broth = 5cP

cake resistance = 1*1011 cm/g

dry basis per volume of filtrate = 20 g/liter

Diameter = 8m ,  Length = 12m

vacuum pressure = 80 kpa

cake formation time = 20 s

cycle time = 60 s

Determine the filtration rate in volumes/hr  expected fir the rotary vacuum filter

attached below is a detailed solution of the question

Hence The filtration rate in volumes/hr expected for the rotary vacuum filter

V' = ( [tex]\frac{60}{20}[/tex] ) * 1706.0670

   = 5118.201 liters  ≈ 5.118 m^3/hr

An isolated pretimed signalized intersection has an approach with a traffic flow rate of 750 veh/h and a saturation flow rate of 3200 veh/h. This approach is allocated 32 seconds of effective green time. The cycle length is 100 seconds. Determine the average approach delay a) 4,6 s b) 30.2 s c) 34.8 s d) 35.0 s​

Answers

Answer: This approach is allocated 32 seconds of effective green time. The cycle length is 100 seconds. Determine the average approach delay (using Eq. 7.27). A) 34.8 s.

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16.5 s later. The descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude. Assume that g = 32.2 ft/s2.
Determine
(a) the speed v1 of the rocket at the end of powered flight,
(b) the maximum altitude reached by the rocket.

Answers

Answer:

[tex]u = 260.22m/s[/tex]

[tex]S_{max} = 1141.07ft[/tex]

Explanation:

Given

[tex]S_0 = 89.6ft[/tex] --- Initial altitude

[tex]S_{16.5} = 0ft[/tex] -- Altitude after 16.5 seconds

[tex]a = -g = -32.2ft/s^2[/tex] --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

[tex]S = ut + \frac{1}{2}at^2[/tex]

The final altitude after 16.5 seconds is represented as:

[tex]S_{16.5} = S_0 + ut + \frac{1}{2}at^2[/tex]

Substitute the following values:

[tex]S_0 = 89.6ft[/tex]       [tex]S_{16.5} = 0ft[/tex]     [tex]a = -g = -32.2ft/s^2[/tex]    and [tex]t = 16.5[/tex]

So, we have:

[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2[/tex]

[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45[/tex]

[tex]0 = 89.6 + 16.5u- 4383.225[/tex]

Collect Like Terms

[tex]16.5u = -89.6 +4383.225[/tex]

[tex]16.5u = 4293.625[/tex]

Make u the subject

[tex]u = \frac{4293.625}{16.5}[/tex]

[tex]u = 260.21969697[/tex]

[tex]u = 260.22m/s[/tex]

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

[tex]v=u + at[/tex]

At the maximum height:

[tex]v =0[/tex] --- The final velocity

[tex]u = 260.22m/s[/tex]

[tex]a = -g = -32.2ft/s^2[/tex]

So, we have:

[tex]0 = 260.22 - 32.2t[/tex]

Collect Like Terms

[tex]32.2t = 260.22[/tex]

Make t the subject

[tex]t = \frac{260.22}{ 32.2}[/tex]

[tex]t = 8.08s[/tex]

The maximum height is then calculated as:

[tex]S_{max} = S_0 + ut + \frac{1}{2}at^2[/tex]

This gives:

[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2[/tex]

[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22[/tex]

[tex]S_{max} = 89.6 + 260.22 * 8.08 - 1051.11[/tex]

[tex]S_{max} = 1141.0676[/tex]

[tex]S_{max} = 1141.07ft[/tex]

Hence, the maximum height is 1141.07ft

3) Explain how dc machines Can work as motor and generator​

Answers

The working principle of a DC machine is when electric current flows through a coil within a magnetic field, and then the magnetic force generates a torque that rotates the dc motor. The DC machines are classified into two types such as DC generator as well as DC motor.

How are new slip rings attached to the rotor core during alternator rebuilding?

Answers

The new slip rings is been  attached to the rotor core during alternator rebuilding by b. Soldered on.

What Soldering ?

Soldering serves as the way of joining  different types of metals together by melting solder.

This is been used on new slip rings to get it  attached to the rotor core during alternator rebuilding .

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CHECK THE COMPLETE QUESTION BELOW:

How are new slip rings attached to the rotor core during alternator rebuilding?

a. Machine clamped

b. Soldered on

c. Crimped on

d. epoxy glue

A 50-mm cube of the graphite fiber reinforced polymer matrix composite material is subjected to 125-kN uniformly distributed compressive force in the direction 2, which is perpendicular to the fiber direction (direction 1). The cube is constrained against expansion in direction 3. Determine:

a. changes in the 50-mm dimensions.
b. stresses required to provide constraints.

Answers

Answer:

hello some parts of your question is missing attached below is the missing part

answer :

A) Determine changes in the 50-mm dimensions

The changes are : 0.006mm compression  in y-direction

                               0.002 mm expansion in x and z directions

B) the stress required are evenly distributed

Explanation:

Given data :

50-mm cube of graphite fiber reinforced polymer matrix

subjected to 125-KN force in direction 2,

direction 2 is perpendicular to fiber direction ( direction 1 ) and cube is constrained against expansion in direction 3

A) Determine changes in the 50-mm dimensions

The changes are : 0.006mm compression  in y-direction

                               0.002 mm expansion in x and z directions

B) the stress required are evenly distributed

attached below is the detailed solution

The demand schedules for Jones, Smith, and other buyers are shown in the table below:
Draw the 4 demand curves for i) Jones ii) Smith iii) Other buyers and iv) All buyers.

Answers

The demand curve is the graphical representation of the relationship between the price of a good and the quantity demanded for a given period of time.

What is a demand schedule?

A demand schedule is a table which shows the quantity demanded of a good or service at different price levels.

A demand schedule can be graphed as a continuous demand curve on a chart where the Y-axis represents the price and the X-axis represents quantity.

Here, a typical representation, the price will appear on the left vertical axis, the quantity demanded on the horizontal axis.

Note that the complete information wasn't found and an overview was given.

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For some alloy, the yield stress is 345-MPa (50,000-psi) and the elastic modulus (E) is 103-GPa (15x106 psi). What is the maximum load that may be applied to a specimen with a cross-sectional area of 130-mm2 (0.2-in2) without plastic deformation

Answers

The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is; 35535 N

How to find Elastic Modulus?

We are told that for an alloy, the yield stress is 345-MPa and the elastic modulus (E) is 103-GPa.

Now, we want to find the maximum load that may be applied to a specimen with a cross-sectional area of 130-mm² without plastic deformation. Thus;

We are given the parameters;

Yield Stress; σ = 345 Mpa = 345 * 10⁶ Pa

Elastic Modulus; E = 103 GPa = 103 * 10⁹ Pa

Cross sectional Area; A = 130 mm² = 103 * 10⁻⁶ m²

Formula for stress without Plastic deformation is;

σ = F_max/Area

where;

σ is stress

F_max is maximum force

Area is Area

Thus making maximum force the subject of the formula gives;

F_max = σ * A

Plugging in the relevant values for stress and area gives us;

F_max = 345 * 10⁶ * 103 * 10⁻⁶

F_max = 35535 N

The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is gotten to be 35535 N

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The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a particle) is 2 m above the road. Calculate the speed v of the truck.

Answers

Answer:

19.81 m/s

Explanation:

The total acceleration of the truck (a) is due to the centripetal acceleration and as a result of the linear acceleration. Therefore the total acceleration (a) is given by:

[tex]a^2=a_n^2+a_t^2\\\\where\ a_n=centripetal\ acceleration=\frac{v^2}{r},a_t=linear \ acceleration\\\\But\ since\ the \ speed\ is \ constant, the \ linear \ acceleration(a_t)\ would\ be\ 0.\\\\a^2=a_n^2+a_t^2\\\\a^2=a_n^2\\\\a=a_n=\frac{v^2}{r} \\\\v^2=ar\\\\v=\sqrt{ar} \\\\a=0.4g=0.4*9.81,r=98\ m+2\ m=100\ m\\\\v=\sqrt{0.4*9.81*100} \\\\v=19.81\ m/s[/tex]

Where is the primary area for romanesque architectural sculpture?

Answers

Answer:

on the façade, entrance, and column capitals

Airtight plugs are installed in the ends of large sizes of PVC conduit during bending to_____________.

Answers

Airtight plugs are installed in the ends of large sizes of PVC conduit before bending to  A. prevent the conduit from collapsing when heated.

What is Airtight plugs?

Airtight plugs is necessary in the  PVC conduit so as to be able to avoid the collapse of conduit.

In this case, Airtight plugs are installed in the ends of large sizes of PVC conduit before bending to  A. prevent the conduit from collapsing when heated.

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who was part of dempwolf his firm when he first started

Answers

Explanation:

Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with  brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.

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