How much energy does it take to boil water for pasta? For a one-pound box of pasta
you would need four quarts of water, which requires 15.8 kJ of energy for every degree
Celsius (°C) of temperature increase. Your thermometer measures the starting
temperature as 48°F. Water boils at 212°F.
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?

Answers

Answer 1

Answer:

a.  164 °F b. 91.11 °C c. 1439.54 kJ

Explanation:

a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?

Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.

So, Δ°F = 212 °F - 48 °F = 164 °F

b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?

To find the degree change in Celsius, we convert the initial and final temperature to Celsius.

°C = 5(°F - 32)/9

So, 48 °F in Celsius is

°C₁ = 5(48 - 32)/9

°C₁ = 5(16)/9

°C₁ = 80/9

°C₁ = 8.89 °C

Also, 212 °F in Celsius is

°C₂ = 5(212 - 32)/9

°C₂ = 5(180)/9

°C₂ = 5(20)

°C₂ = 100 °C

So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.

So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C

c. [2 pts] How much energy is required to heat the four quarts of water from

48°F to 212°F (boiling)?

Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise =  15.8 kJ/°C × 91.11 °C = 1439.54 kJ


Related Questions

The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the outside surface of the solid shaft. Determine the principal stresses, the maximum in-plane shear stress and average normal stress using Mohr's Circle.

Answers

Answer: Attached below is the missing detail and Mohr's circle.

i) б1 =  9.6 Ksi

б2 = -10.7 ksi

ii) 10.2 Ksi

iii)  -0.51Ksi

Explanation:

First step :

direct compressive stress on shaft

бd = P / π/4 * d^2

      = -20 / 0.785 * 5^2  = -1.09 Ksi

shear stress at the outer surface due to torsion

ζ = 16*T / πd^3

  = (16 * 250 ) / π * 5^3  = 010.19 Ksi

Calculate the Principal stress, maximum in-plane shear stress and average normal stress

Using Mohr's circle ( attached below )

i) principal stresses:

б1 = 4.8 cm * 2 = 9.6 Ksi

б2 = -5.35 cm * 2 = -10.7 ksi

ii) maximum in-plane shear stress

ζ  = radius of Mohr's circle

   = 5.1 cm = 10.2 Ksi   ( Given that ; 1 cm = 2Ksi )

iii) average normal stress

 = 9.6 + ( - 10.7 ) / 2

  = -0.51Ksi

A micromechanical resonator is to be designed to have a Q factor of 1000 and a natural frequency of 2 kHz. Determine the system-damping factor and the system bandwidth.

Answers

Answer:

Explanation:

Given:

Q factor, =1000

natural frequency, [tex]f_n=2000~Hz[/tex]

Damping factor, [tex]\zeta=?[/tex]

Bandwidth, BW=?

We have the relation:

[tex]Q=\frac{1}{2\zeta}[/tex]

[tex]\zeta=\frac{1}{2Q}[/tex]

[tex]\zeta=\frac{1}{2\times 1000}[/tex]

[tex]\zeta=5\times 10^{-4}[/tex]

Bandwidth:

[tex]BW=\frac{f_n}{Q}[/tex]

[tex]BW=\frac{2000}{1000}[/tex]

[tex]BW=2~Hz[/tex]

Atmospheric pressure is 101 kPa. Pressure inside a tire is measured using a typical tire pressure gage to be 900 kPa. Find gage pressure and absolute pressure in the tire. ___________________________________________________________________

Answers

Answer:

The gage and absolute pressures are 900 and 1001 kilopascals, respectively.

Explanation:

The gage pressure ([tex]P_{g}[/tex]), in kilopascals, is the difference between absolute ([tex]P_{abs}[/tex]) and atmospheric pressures ([tex]P_{atm}[/tex]), measured in kilopascals. If we know that [tex]P_{g} = 900\,kPa[/tex] and [tex]P_{atm} = 101\,kPa[/tex], then the gage and absolute pressures are, respectively:

[tex]P_{g} = 900\,kPa[/tex]

[tex]P_{abs} = P_{atm} + P_{g}[/tex]

[tex]P_{abs} = 101\,kPa + 900\,kPa[/tex]

[tex]P_{abs} = 1001\,kPa[/tex]

The gage and absolute pressures are 900 and 1001 kilopascals, respectively.

what type of slab and beam used in construction of space neddle​

Answers

jrjrkeekkekekkwkkakkllalllalallalllalalaallalalaalalalalalallallallllallalalallaaallalallllllallllllllalaalalalaaaaalalaaaaaaalgjgiejxpwunfifjruritiririirieoeowowowowowowowowooeowowowoeeoeowowowowowowowoowowwowowowoozoeisiaokseekxidjdkdjfidjfjdjfjfjrifjrifjdirjdjrjfjrjfjrjfjrfuejwwuxmaneanfjkaosndjxieneamalhaqzeeshanvhorahfuensiwjakaksjdhfhfnfhfndjxnxmakaalalalwlwlwwow

Future solution for air pollution in new zealand

Answers

Answer:

New Zealand may use some of these solutions to prevent air pollution

Explanation:

Using public transports.

Recycle and Reuse

No to plastic bags

Reduction of forest fires and smoking

Use of fans instead of Air Conditioner

Use filters for chimneys

Avoid usage of crackers

How do your arm muscles work to lift a mug of coffee to your lips?
A.
The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.
B.
The muscle tendons in your arms stretch to their limit as a result of the flexibility exercises that lift the mug.
C.
Your muscles rapidly increase the speed at which they twitch to trigger your arm movement.
D.
Your biceps and triceps muscles first relax, and then the arm muscle raises the mug.

Answers

Answer:

A. The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.

Explanation:

Skeletal muscle is also known as voluntary muscle and it can be defined as a type of muscle connected with the skeleton by tendons, so as to form a mechanical system that enables the movement of the limbs and other body parts with respect to another.

Generally, skeletal muscles are only found in vertebrates.

The muscles in the arm work to lift a mug of coffee to your lips when the biceps muscle contracts while the triceps muscle relaxes.

Basically, the only way the forearm move up to lift a mug for example, is through the contraction of the biceps muscle and the relaxation of the triceps muscle.

This ultimately implies that, the elbow joint is straightened or bent due to the actions of the biceps muscle and triceps muscle against each other.

To bend the elbow and raise the forearm, the biceps muscle contracts and the triceps muscle relaxes.To straighten the elbow and lower the forearm, the biceps muscle relaxes and the triceps muscle contracts.

Answer:A

Explanation:

got it right on plato

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution.
x dx/dy−y=x^2sinx

Answers

Answer:

Interval:  x∈ ( 0, ∞ )

There are no transient terms

Explanation:

x (dy/dx) – y= x^2sinx

Attached below is the detailed solution of the Given problem

There are no transient terms found in the general solution

Interval:  x∈ ( 0, ∞ )

In a CNC machining operation, the has to be moved from point (5, 4) to point(7, 2)along a circular path with center at (7,2). Before starting operation, the tool is at (5, 4).The correct G and M code for this motion is

Answers

Answer: hello your question is incomplete below is the complete question

answer:

N010 GO2 X7.0 Y2.0 15.0 J2.0  ( option 1 )

Explanation:

Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path

GO2 = circular interpolation in a clockwise path

G91 = incremental dimension

hence the correct option is :

N010 GO2 X7.0 Y2.0 15.0 J2.0  

An intersection with a four phase signal has a displayed red time of 35 seconds, a start-up lost time of 2 seconds, a yellow time of 4 seconds, and an all red time of 1 second per phase. The total lost time is typically calculated as ____ seconds per cycle.

Answers

Answer:

53 sec / cycle

Explanation:

Displayed red time = 35 seconds

Start up lost time = 2 seconds

Yellow time = 4 seconds

Red time = 1 second

Total lost time L = 2n + r

L = lost time

n = number of phase

R = red time

35+2+4+4*1

= 45

L = 2x4+45

= 53 sec/cycle

The total lost time is typically calculated as 53 seconds per cycle

where are the field poles mounted on an alternator

Answers

Answer:

The magnetic field for this type of alternator is established by a set of stationary field poles mounted on the periphery of the alternator frame. The field flux created by these poles is cut by conductors inserted in slots on the surface of the rotating armature.

Determine the transfer function H(s) for a high pass filter with the following characteristics:
1. a cutoff frequency of 100 kHz
2. a stopband attenuation rate of 40 dB/decade
3. a nominal passband gain of 20 dB, which drops to 14 dB at the cutoff frequency
Write the formula for H(s) that satisfies these requirements.

Answers

Answer:

H(s) =  10 / [ 1 + s / (200*10^3π ) ]^2

Explanation:

Characteristics of the high pass filter

Cutoff frequency = 100 kHz

stopband attenuation rate = 40 dB/decade

nominal passband gain = 20dB  = 20logK = 20

Formula for H(s) satisfying the requirements above

given that the stopband attenuation = 40 dB/decade the formula for H(s) that will satisfy the requirements is a second order filter

H(s) = K / ( 1 + s/Wo ) ^2  ----- ( 1 )

Wo = 2πf = 2π ( 100 * 10^3 ) =  200 * 10^3 π

 K = 10

back to equation ( 1 )

H(s) =  10 / [ 1 + s / (200*10^3π ) ]^2

In low speed subsonic wind tunnels, the value of test section velocity can be controlled by adjusting the pressure difference between the inlet and test-section for a fixed ratio of inlet-to-test section cross-sectional area.
a. True
b. false

Answers

Answer:

Hence the given statement is false.

Explanation:

For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.

The speed within the test section is decided by the planning of the tunnel.  

Thus by adjusting the pressure difference won't change the worth of test section velocity.

Answer:

The given statement is false .

Your shifts productivity is Slow because one person is not pulling his share. The rest of the team is Getting upset.

Answers

Answer:

you are right but then you ddnt ask a question

A designer needs to select the material for a plate under tensile stress. Assuming that the applied tensile force is 13,000 lb and the area under the stress is 4 square inches, determine which material should be selected to assure safety. Assume safety factor is 2. Material A: Ultimate Tensile stress is 8000 lb/in2Material B: Ultimate Tensile stress is 5500 lb/in2

Answers

Wow Muy bien no se que dice pero bien.

What is the built-in pollution control system in an incinerator called

Answers

Explanation:

hbyndbnn☝️

Reinforced concrete is a raw material that has always been available, but it was unappreciated by architects until the nineteenth century.

a. True
b. False

Answers

Answer: False

Explanation:

Reinforced concrete is simply a combination of the traditional cement concrete with the steel bars which are the reinforcements.

Reinforced concrete is utilized for construction purpose mostly on a large scale. The reinforced concrete was invented by French gardener Joseph in 1849 therefore, it has always been available and appreciated by architects before the 19th century.

A venturimeter of 400 mm × 200 mm is provided in a vertical pipeline carrying oil of specific gravity 0.82, flow being upward. The difference in elevation of the throat section and entrance section of the venturimeter is 300 mm. The differential U-tube mercury manometer shows a gauge deflection of 300 mm. Calculate: (i) The discharge of oil, and (ii) The pressure difference between the entrance section and the throat section.Take the coefficient of meter as 0.98 and specific gravity of mercury as 13.6

Answers

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Where,

is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ =  

A₂ = Area at the throat

A₂ =  

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

or

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Hope This Helps :D

HELP PLEASE!! ASAP!!!!
can some answer this 2 questions please as paragraph i want it nowww it is graded what action should be taken to make it safe ? also the first question

Answers

Actions violated:

Long hair isn't tied upThe girl isn't wearing a lab coatThe girl isn't wearing safety gogglesExtra: There doesn't seem to be an emergency fire blanket in the safe

Actions to be taken:

Make sure the girl wears a lab coat or kick her outMake sure the girl wears safety goggles or kick her outMake sure her hair is tied up or kick her out

Edit: Use these to write your paragraph.

Jodi hasn’t tied her hair up. Jodi is not wearing goggles and Kimberley and Jodie are not wearing gloves

In a true Brayton cycle, the pressure ratio is 9. Air input temperature to the cycle 300 K pressure is 100 kPa. The maximum temperature in the cycle is 1300 K. Compressor and turbine their yields are equal to each other. Net work obtained from the cycle is 225 kJ / kg. Accordingly, the cycle find the overall yield. The specific temperatures are variable.

Answers

Answer:

i did not known answer but anobody help you

A 1m3 tank containing air at 25℃ and 500kPa is connected through a valve to
another tank containing 5kg of air at 35℃ and 200kPa. Now the valve is opened,
and the entire system is allowed to reach thermal equilibrium, which is at 20℃
(Take: Ru = 8.314 kJ / kg.K).

Answers

Answer:

The right answer is "2.2099 m³".

Explanation:

Given:

Mass,

m = 5 kg

Temperature,

T = 35℃

or,

  = 35 + 273

Pressure,

P = 200 kPa

Gas constant,

R = 0.2870 kj/kgK

By using the ideal gas equation,

The volume will be:

⇒ [tex]PV=mRT[/tex]

or,

⇒    [tex]V=\frac{mRT}{P}[/tex]

By substituting the values, we get

          [tex]=\frac{5(0.2870)(35+273)}{200}[/tex]

          [tex]=\frac{441.98}{200}[/tex]

          [tex]=2.2099 \ m^3[/tex]

Cho thanh có tiết diện thay đổi chịu tải trọng dọc trục (hình 1).
Biết d1 = 5 cm, d2 = 8 cm, a= 15 cm, b=10cm, P1 =400kN, P2 =200kN, E= 2.104 kN/cm2.
a) Vẽ biểu đồ lực dọc.
b) Kiểm tra bền của thanh AC, [ϭ] =10 (kN/cm2).
c) Xác định chuyển vị theo phương dọc trục của tâm tiết diện C

Answers

Answer:

saay in English language

You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are currently reviewing the scenario involving the failure of a nitrogen regulator that provides inert padding to the vapor space of the reactor. Your calculations show that the maximum discharge rate of nitrogen through the existing relief system of the vessel is 0.5 kgls, However, your calculations also show that the flow of nitrogen through the l-in supply pipe will be much greater than this. Thus under the current configuration a failure of the nitrogen regulator will result in an over pressuring of the reactor. One way to solve the problem is to install an orifice plate in the nitrogen line, thus limiting the flow to the maximum of 0.5 kg/s. Determine the orifice diameter (in cm) required to achieve this flow. Assume a nitrogen source supply pressure of 15 bar absolute. The ambient temperature is 25°C and the ambient pressure is 1 atm. 3.

Answers

Answer:

[tex]D=0.016m[/tex]

Explanation:

From the question we are told that:

Discharge Rate [tex]F_r=0.5kgls[/tex]

Pressure [tex]P=15Kpa[/tex]

Temperature [tex]T=25=>298K[/tex]

Ambient pressure is 1 atm.

Generally the equation for Density is mathematically given by

[tex]\rho=\frac{PM}{RT}[/tex]

[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]

[tex]\rho=16.958kg/m^2[/tex]

Generally the equation for Flow rate is mathematically given by

[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]

Where

[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]

[tex]Q=1.4[/tex]

[tex]\mu= Discharge\ coefficient[/tex]

[tex]\mu=0.68[/tex]

Therefore

[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]

[tex]A=2.129*10^{-4}[/tex]

Where

[tex]A=\frac{\pi}{4}D^2[/tex]

[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]

[tex]D=0.016m[/tex]

A 20-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15°, what must be the length of the bar?

Answers

Answer:

1.887 m

Explanation:

(15 *pi)/180

= 0.2618 rad

Polar moment

= Pi*d⁴/32

= (22/7*20⁴)/32

= 15707.96

Torque on shaft

= ((22/7)*20³*110)/16

= 172857.14

= 172.8nm

Shear modulus

G = 79.3

L = Gjθ/T

= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8

= 1.887 m

The length of the bar is therefore 1.887 meters

Explain why veracity, value, and visualization can also be said to apply to relational databases as well as Big Data.

Answers

Answer:

Veracity, Value and Visualization are not only the characteristics of Big Data but are also the characteristics of relational databases. Veracity of data is issue with smallest data stores this is the reason that it is important in relation...

Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g

Answers

Answer:

[tex]D=0.41m[/tex]

Explanation:

From the question we are told that:

Discharge rate [tex]V_r=0.35 m3/s[/tex]

Distance [tex]d=4km[/tex]

Elevation of the pumping station [tex]h_p= 140 m[/tex]

Elevation of the Exit point [tex]h_e= 150 m[/tex]

Generally the Steady Flow Energy Equation SFEE is mathematically given by

[tex]h_p=h_e+h[/tex]

With

[tex]P_1-P_2[/tex]

And

[tex]V_1=V-2[/tex]

Therefore

[tex]h=140-150[/tex]

[tex]h=10[/tex]

Generally h is give as

[tex]h=\frac{0.5LV^2}{2gD}[/tex]

[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

Therefore

[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]

[tex]D=0.41m[/tex]

Represent each of the following units as a combination of primitive
dimensions where M=mass, L=length, T=time. As an example, miles per hour would
correspond to [L/T].

a. kilometer
b. quart
c. pascal
d. watt
e. newton
f. horsepower

Answers

Answer:

a. unit of length: [L]

b. unit of volume: [[tex]L^3[/tex]]

c. unit of pressure:[tex]P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]}[/tex] [tex][ML^{-1}T^{-2}][/tex]

d. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]

e. unit of force: [tex][kg.m/s^2]\equiv [MLT^{-2}][/tex]

f. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]

Force: [tex]F=m.a=m.\frac{v}{t}=m.\frac{x}{t}\div t[/tex]

Power: [tex]P=\frac{W}{t}=\frac{F.x}{t}[/tex]

where:

F = force

A = area

W = work

t = time

a = acceleration

v = velocity

x = displacement

A stream of oxygen enters a compressor at a rate of 200 SCMH. The oxygen exits at 360 K and 500 bar. Determine the volumetric flowrate exiting the compressor using the compressibility factor equation of state.

Answers

Answer:

≈ 0.516 m^3/hr

Explanation:

Inlet of compressor = 200 SCMH

sheer standard conditions = 1 atm and 288.5 K

For oxygen :

critical pressure(Pc) = 49.8 atm

critical temperature Tc = 154.6 K

hence at compressor inlet

Tr = T / Tc = 288.5/154.6 = 1.866

Pr = P / Pc = 1 / 49.8 = 0.0204

Z1 ( from compressibility chart ) = 0.98

at compressor outlet

P2 = 500 bar = 500*0.9869 = 493.45 atm  , T2 = 360 k

hence : Pr = P / Pc = 493.45 / 49.8 = 9.91

            Tr = T / Tc = 360 / 154.6 = 2.33

Z2 ( from compressibility chart ) ≈ 1

V2( volumetric flow rate ) = V1*(P₁Z₂T₂) / (P₂Z₁T₁)

                                         = 200 ( 1 * 1* 360) / (493.45 *0.98*288.5)

                                         = 0.516 m^3/hr

Using 1.5 V batteries, a switch, and three lamps, devise a circuit to apply 4.5 V across eitherone lamp, two lamps in series, or three lamps in series with a single-control switch. Draw theschematic.

Answers

Answer: the attached picture is the answer.

Explanation:

Assuming:

the switch position connect to 1, hence 4.5V exist at across lamp1

the switch position connects to 2 hence 4.5 V exist across lamp 1 and lamp 2

the switch position connects to 3, hence, 4.5 V exist across lamp 1, lamp 2 and lamp 3.

WILL MARK BRAINLIST I need help on this asap thanks
Determine the dimensions for T if T = M V^2 A / L^3 where M is a mass, V is a velocity, A is an area, and L is a length.


L / T


M


M L / T^2


M / (L T^2)


No dimensions

Answers

Explanation:

ask your dad please and use your brain

All of the following are instruments involved in changing a tire EXCEPT:

Answers

Answer:

can you please give us the option for this question

What are the other options mrs.gurl (I call everyone that not trying 2 be rude)
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