In a production facility, 1.6-in-thick 2-ft × 2-ft square brass plates (rho = 532.5 lbm/ft3 and cp = 0.091 Btu/lbm·°F) that are initially at a uniform temperature of 75°F are heated by passing them through an oven at 1500°F at a rate of 340 per minute. If the plates remain in the oven until their average temperature rises to 900°F, determine the rate of heat transfer to the plates in the furnace.

The net work done during the Stirling cycle is 2.7 kJ.

In order to determine the net work done during the Stirling cycle, we first need to calculate the initial and final volumes of the air in the piston-cylinder assembly.

Given:

The compression ratio is given by:

r = V1/V2

where V2 is the final volume. Solving for V2, we get:

V2 = V1/r = 0.03/7.5 = 0.004 m3

Next, we can use the ideal gas law to relate the initial and final pressures and volumes:

P1V1/T1 = P2V2/T2

Since the process is isothermal, T1 = T2 = TH = 1200 K.

Solving for P2, we get:

P2 = (P1V1T2)/V2T1 = (100,000 x 0.03 x 1200)/(0.004 x 1200) = 100,000 Pa

Finally, we can calculate the net work done using the formula:

Wnet = (P1V1) - (P2V2)

Substituting the values, we get:

Wnet = (100,000 x 0.03) - (100,000 x 0.004) = 2700 J = 2.7 kJ

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

Answer:

The solution is given in the attachments.

Answer:

The detailed answer to the question is explained in the attached file.

Explanation:

Answer:

The data file with 1000 integers

for merge sort the time efficiency is (1000×㏒1000) = 1000 × 3 = 3000

for insertion sort the time efficiency is (1000 × 2) = 2000

The data file with 1,000,000 integers

for merge sort the time efficiency is (1,000,000×㏒1,000,000) = 1,000,000 × 6 = 6,000,000

for insertion sort the time efficiency is (1,000,000 × 2) = 2,000,000

Explanation:

The execution time or temporal complexity refers to how the execution time of an algorithm increases as the size of input increases

For merge sort, the time efficiency  is given by the formula

For insertion sort the time efficiency  is given by the formula

Where n refers to the size of input

Answer: For the center plate to remain stationed in one position without rotating, the bottom plate has to move to the left at a speed of 2m/s, so as to cancel the force acting on it from the top.

The center plate will not move when the bottom plate is moving left in a speed of 2m/s to counter the speed of the top plate, because a body will continue to be at rest if all the forces acting towards the body are equal. The center plate will be at rest because we have directed equal force from the top and bottom of the plate.

Answer:

221.929 hp

98.19%

Explanation:

See attached picture.

Answer:

mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h

Explanation:

The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h

The rate of evaporation will be "2.871 Kg/hour".

According to the question,

Rate of heat supplied:

= 60% of 3 kW

= 1.8 kW

= 1.8 KJ/s

Vaporization of water, [tex]\Delta H = 2257 \ KJ/Kg[/tex]

Time taken will be:

= [tex]\frac{2257}{1.8}[/tex]

= [tex]1254 \ s[/tex]

= [tex]\frac{1254}{3600} \ hour[/tex]

= [tex]0.3482 \ hour[/tex]

hence,

The rate of evaporation,

= [tex]\frac{Mass \ of \ water}{Time}[/tex]

= [tex]\frac{1}{0.3482}[/tex]

= [tex]2.871 \ Kg/hour[/tex]

Thus the above answer is right.

Find out more information about evaporation here:

https://brainly.com/question/4406110

Answer:

a) 500 rev/min.

b) 50 Hz.

Explanation:

See the attached pictures.

Answer:

Quality strategy is an integral part of organization's strategy that ensures quality. It involves market and productivity strategies that have a very high significance.  A successful quality strategy must endeavour to have features such as; engaging employees in the necessary activities to implement quality, an organizational culture that fosters quality and an understanding of the principles of quality.

All these will enable the organization to stand out and beat competition.

A successful quality strategy features includes engaging the employees:

A quality strategy serve as an integral part of organization's strategy that ensures quality.

The quality strategy involves market and productivity strategies that have a very high significance.

Hence, the successful quality strategy features includes engaging the employees in necessary activities to implement quality, organizational culture that fosters quality and understanding the principles of quality.

Therefore, all the option is correct.

Read more about quality strategy

brainly.com/question/24462624

Answer:

a) [tex]Re_{D} = 111896.745[/tex], b) [tex]Re_{D} = 1.119\times 10^{-7}[/tex]

Explanation:

a) The Reynolds number for the water flowing in a circular tube is:

[tex]Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}[/tex]

Let assume that density and dynamic viscosity at 25 °C are [tex]997\,\frac{kg}{m^{3}}[/tex] [tex]0.891\times 10^{-3}\,\frac{kg}{m\cdot s}[/tex], respectively. Then:

[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]

[tex]Re_{D} = 111896.745[/tex]

b) The result is:

[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]

[tex]Re_{D} = 1.119\times 10^{-7}[/tex]

Answer:

See explanation

Explanation:

Airflow t

T¥, h D

Ts

w

Heater

h

(q•, k ) L

(a) Under conditions for which a uniform surface temperature Ts is maintained around the circum- ference of the heater and the temperature Too and convection coefficient h of the airflow are known, obtain an expression for the rate of heat transfer per unit length to the air. Evaluate the

heat rate for Ts = 300°C, D = 20 mm, an alu- minum sleeve (ks = 240 W/m · K), w = 40 mm, N = 16, t = 4 mm, L = 20 mm, Too = 50°C, and h = 500 W/m2 · K.

Answer:

Air void content of core 2 = 5.87%

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Full Question

1. Correct the following code and

2. Convert the do while loop the following code to a while loop

declare integer product

declare integer number

product = 0

do while product < 100

display ""Type your number""

input number

product = number * 10

loop

display product

End While

Answer:

1. Code Correction

The errors in the code segment are:

a. The use of do while on line 4

You either use do or while product < 100

b. The use of double "" as open and end quotes for the string literal on line 5

c. The use of "loop" statement on line 7

The correction of the code segment is as follows:

declare integer product

declare integer number

product = 0

while product < 100

display "Type your number"

input number

product = number * 10

display product

End While

2. The same code segment using a do-while statement

declare integer product

declare integer number

product = 0

Do

display "Type your number"

input number

product = number * 10

display product

while product < 100

Answer:

I = 14.44A

Explanation:

calculating 80% of the circuit breaker current

I = (80/100)20A = 16A

eight 500W lamps are to be installed on the circuit. assume they are connected in parallel

total power = 500 x 8 = 4000W

power = voltage x current

current = power/voltage = 4000W/277V = 14.4A

The current obtained is less than 80% of the circuit breaker.

20A circuit breaker is large enough to carry the load.

Answer:

attached below

Explanation:

Answer:

FALSE.

Explanation:

This statement can be considered false, because it is important to know essential aspects of the prospect before starting the sales dialogue.

In addition to obtaining basic information about the client, it is relevant for the company to know the prospect's organization.

We can define the prospect as the one who, for the company, has the ideal customer profile, but who has not yet shown interest in consuming its products and services.

Therefore, knowing information about it, about its basic and complex characteristics will help the organization to develop sales and marketing strategies aimed at attracting the prospect.

Answer:

P = 188.496 N.

Explanation:

The axial elongation of the wire is modelled by this equation:

[tex]\delta = \frac{P \cdot L}{(\frac{\pi}{4}\cdot D^{2})\cdot E_{steel}}[/tex]

The force can be calculated by isolating it within the expression:

[tex]P=\frac{\delta \cdot (\frac{\pi}{4}\cdot D^{2})\cdot E_{steel}}{L}[/tex]

Let assume that [tex]L = 1 m[/tex]. Then:

[tex]P =\frac{(0.3 mm)\cdot (\frac{1 m}{1000 mm} )\cdot (\frac{\pi}{4}\cdot [(2 mm)\cdot(\frac{1 m}{1000 mm} )]^{2} )\cdot (200 \times 10^{9} Pa)}{1 m} \\P \approx 188.496 N[/tex]

Answer:

Negative feedback

Explanation:

In Biology, negative feedback refers to the counteraction of an effect by its own influence on the process producing it. For instance, the presence of   a high level of a particular hormone in the blood may inhibit further secretion of that hormone.

In other words, in negative feedback, the result of a certain action may inhibit further performance of that action

Answer:

A) The quality of the refrigerant at the evaporator inlet = 0.48

B) The refrigeration load = 5.39 kW

C) COP = 2.14

Explanation:

A) From the refrigerant R-144 table I attached,

At P=60kpa and interpolating at - 34°C,we obtain enthalpy;

h1 = 230.03 Kj/kg

Also at P= 1.2MPa which is 1200kpa and interpolating at 65°C,we obtain enthalpy ;

h2 = 295.16 Kj/Kg

Also at P= 1.2MPa which is 1200kpa and interpolating at 42°C,we obtain enthalpy ;

h3 = 111.23 Kj/Kg

h4 is equal to h3 and thus h4 = 111.23 Kj/kg

We want to find the refrigerant quality at the evaporation inlet which is state 4 and P= 60 Kpa.

Thus, from the table attached, we see that hf = 3.84 at that pressure and hg = 227.8

Now, to find the quality of the refrigerant, we'll use the formula,

x4 = (h4 - hg) /(hf - hg)

Where x4 is the quality of the refrigerant. Thus;

x4 = (111.23 - 3.84)/(227.8 - 3.84) = 0.48

B) The mass flow rate of the refrigerant can be determined by applying a 1st law energy balance across the

condenser. Thus, the water properties can be obtained by using a saturated liquid at the given temperatures;

So using the first table in the image i attached; interpolating at 18°C; hw1 = hf = 75.54 kJ/kg

Also interpolating at 26°C; hw2 = hf = 109.01 kJ/kg

Now;

(mr) (h2 − h3)= (mw) (hw2 − hw1)

mr is mass flow rate

Making mr the subject, we get;

mr = [(mw) (hw2 − hw1)] /(h2 − h3)

mr = [(0.25 kg/s)(109.01 − 75.54) kJ/kg

] /(295.13 − 111.37) kJ/kg

mr = 8.3675/183.76

mr = 0.0455 kg/s

Formula for refrigeration load is;

QL = mr(h1 − h4)

Thus,

QL = (0.0455 kg/s)(230.03 − 111.37) kJ/kg = 5.39 kW

C) The formula for specific work into the compressor is;

W(in) = [(h2 − h1)] − (Q(in

)/mr)

= (295.13 − 230.03) kJ/kg − (0.450kJ/s

/0.0455 kg/s)

= 65.10 Kj/kg - 9.89 Kj/kg

55.21 kJ/kg

Formula for COP is;

COP = qL

/W(in)

Thus; COP = (h1 − h4)/W(in

)

= [(230.03 − 111.37) kJ/kg

] /55.24 kJ/kg

= 2.14

Answer:

a) 5 V.

b) Energy also become half.

Explanation:

See attached picture.

Answer:

the change in length of the elastic rod is 0.147 mm and the vertical displacement of the point b is 0.1911 mm.

Explanation:

As the complete question is not visible , the complete question along with the diagram is found online and is attached herewith.

Part a

Take moments at point A to calculate the tensile force on the steel

rod at point C

[tex]\sum M_A=0\\T_c*2.5-P(2.5+0.75)-q*2.5*2.5/2=0\\T_c*2.5-10(3.25)-5*3.125=0\\T_c=19.25 kN\\[/tex]

Calculate the change in length of the elastic rod

[tex]\Delta l=\dfrac{T_c*L}{A*E}\\\Delta l=\dfrac{19.25\times 10^3*0.75}{\pi/4\times 0.025^2*200\times 10^9}\\\Delta l=0.000147 m \approx 0.147 mm[/tex]

So the change in length of the elastic rod is 0.147 mm.

As by the relation of the similar angles the vertical displacement of the point B is given as

[tex]\Delta B=\Delta l\dfrac{AC+BC}{AC}[/tex]

Here AC=2.5 m

BC=0.75 m

Δl=0.147 mm so the value is as

[tex]\Delta B=\Delta l\dfrac{AC+BC}{AC}\\\Delta B=0.147 mm\dfrac{2.5+0.75}{2.5}\\\Delta B=0.147 mm\dfrac{3.25}{2.5}\\\Delta B=0.1911 mm[/tex]

So the vertical displacement of the point b is 0.1911 mm.

Answer:

[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

[tex]\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}[/tex]

The initial pressure is:

[tex]P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}[/tex]

The speed at outlet is:

[tex]v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}[/tex]

[tex]v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }[/tex]

[tex]v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )[/tex]

The initial pressure is:

[tex]P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}[/tex]

[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]

Answer:

P1 = 42.93 psi

Explanation:

For incompressible fluid, we know that;

A1V1 = A2V2

Making V1 the subject, we obtain;

V1 = A2V2/A1

Now A2V2 is the volumetric flow rate (V') .

Thus; V1 = V'/A1

A1 = πD²/4

Thus, V1 = 4V'/πD²

V' = 250 gal/min

But the diameter is in inches, let's convert to inches³/seconds.

Thus, V' = 250 x 3.85 = 962.5 in³/s

Substituting the relevant values to obtain,

V1 = (4 x 962.5)/(π x 3²) = 136.166 in/s.

Now let's convert to ft/s;

V1 = 136.166 x 0.0833 = 11.34 ft/s

Also for V2;

V2 = (4 x 962.5)/(π x 1.125²) = 968.29 in/s.

Now let's convert to ft/s;

V2 = 968.29 x 0.0833 = 80.66 ft/s

Setting bernoulli equation between the hose and the exit, we obtain;

(p1/γ) + (V1²/2g) = V2²/2g

Where V1 and V2 are intial and final velocities and γ is specific weight of water which is 62.43 lb/ft³ and g i acceleration due to gravity which is 32.2 ft/s²

Making p1 the subject, we obtain;

p1 = (γ/2g)(V2² - V1²)

p1 = (62.43/(2x32.2))(80.66² - 11.34²)

p1 = 6182.35 lb/ft²

So Converting to psi, we have;

p1 = 6182.35/144 = 42.93 psi

Answer:

Hello there, see step by step explanation for answers

Explanation:

A database design for a University Registrar. The following requirements are for designing a Database Schema.

It should include:

1. Information about Students .

2. Information about Departments.

3. Information about Professors.

4. Information about courses.

5 . student Grades.

6. TA's for a course.

7. Department offering different courses.

For Designing Database for Registrar System We need to define various

1. Entity: Student, Course, Instructor, Course offering.

2. Attributes of Entities:

(a) Student entity has Sid, name, program as its attributes

(b)Course has Course_n, title, credits, and syllabus as its attributes

(c) instructor has iid ,name ,dept, title as its attributes

(d) course offering has section_no ,time, room, year ,semester as its attributes

3. Relationship among various entities

(a) enrolls

(b) teaches

© is offered

E-R Diagram for University Registrar

This Diagram shows student entity enrolls various courses which can be having teaches relationship with instructor. Course is being offered by relationship is offered by course offering .

A database design for a University Registrar. The following requirements are for designing a Database Schema.

It should include:

1. Information about Students .

2. Information about Departments.

3. Information about Professors.

4. Information about courses.

5 . student Grades.

6. TA's for a course.

7. Department offering different courses.

For Designing Database for Registrar System We need to define various

1. Entity: Student, Course, Instructor, Course offering.

2. Attributes of Entities:

(a) Student entity has Sid, name, program as its attributes

(b)Course has Course_n, title, credits, and syllabus as its attributes

(c) instructor has iid ,name ,dept, title as its attributes

(d) course offering has section_no ,time, room, year ,semester as its attributes

3. Relationship among various entities

(a) enrolls

(b) teaches

© is offered

E-R Diagram for University Registrar

This Diagram shows student entity enrolls various courses which can be having teaches relationship with instructor. Course is being offered by relationship is offered by course offering .

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

Answer:

1.10 %

Explanation:

The enrichment of the product can be calculated using the following equation:

[tex] \frac{W}{P} = \frac{x_{p} - x_{f}}{x_{f} - x_{w}} [/tex]   (1)

where W: is waste or tails = 26000, P: is the product = 32000, xp: is the wt. fraction of ²³⁵U in product, xf: is the wt. fraction of ²³⁵U in feed = 0.72% and xw: is the wt. fraction of ²³⁵U in waste or tails = 0.25%.      

By solving equation (1) for xp, we can find the enrichment of the product:

[tex] x_{p} = \frac {W}{P} \cdot (x_{f} - x_{w}) + x_{f} [/tex]        

[tex] x_{p} = \frac {26000}{32000} \cdot (0.72 - 0.25) + 0.72 = 1.10% [/tex]  

Therefore, the enrichment of the product is 1.10 %.

I hope it helps you!

Without specific property data for refrigerant-134a at the given conditions, the final temperature and change in specific internal energy cannot be calculated directly. Typically, such tasks require referencing property tables or using equations of state for refrigerant-134a.

The question concerns the process of compressing a refrigerant-134a in a piston-cylinder device while keeping the pressure constant. Given that 100 kg of refrigerant-134a at 200 kPa is initially contained within a volume of 12.322 m3 and then is compressed to half its volume without changing the pressure, we are tasked with determining the final temperature and the change in specific internal energy of the refrigerant.

Since the pressure remains constant during the compression, this scenario can be classified as an isobaric process. However, without specific data such as the initial temperature or specific internal energy values for refrigerant-134a at the given states, we cannot directly calculate the final temperature and change in specific internal energy. Typically, these calculations would require consulting refrigerant-134a property tables or equations of state appropriate for refrigerant-134a to find values at the specified initial and final volumes while maintaining constant pressure.

To find the final temperature, one would typically use the relationship between pressure, volume, and temperature given by the ideal gas law or the specific equations pertaining to refrigerant-134a. The change in specific internal energy could then be determined using specific heat capacities if assuming an ideal gas behavior, or more accurately, through the refrigerant-134a property tables looking at the specific internal energy values at the initial and final states.

Answer:

34.06 W.

Explanation:

Assumptions : ideal turbine " no loss of work", no pipe or friction losses, all the available energy of water in converted to useful power.

(A) total volume of water per day in cubic meters:

1 cubic meters=1000 L

average flow from each apartment*total apartments/1000= (100/1000)*100

                                                                                                =10 m^3

total mass of water m= density*volume=1000*10=10000 kg

total energy of water at 30 m height= m*g*h= 10000*9.81*30=2943000 J

if all the available energy is converted to power.

power produced per day=total energy of water / time

time in seconds=24*3600=86400 s

power produced in a day=2943000/86400= 34.06 W

Answer:

34.06 W

Explanation:

Assuming ideal conditions which are assuming no fiction or pipe loss is made along the line of extraction of water

Energy of water at ideal condition ( Eₐ ) = 1000 kg/m³

height given = 30 meters

quantity of water = 100 Liters

to calculate the quantity of water in M³/s ( cubic per second )

= [tex]\frac{100*10^{-3} }{24*3600}[/tex] = 1.157 * [tex]10^{-6}[/tex]

power produced by water ( Pw) = energy of water * quantity of water in m^3/s

  = 1.157 * [tex]10^{-6}[/tex]  * [tex]10^{3}[/tex] = 1.157 *[tex]10^{-3}[/tex]

since it is an ideal condition all the power produced by water is converted to power produced by the contraception

Pc = ( Pw * g * h ) * n

H = height = 30

g = 9.81

n = number of apartments = 100

Pc =( 1.157 * [tex]10^{-3}[/tex]  * 9.81 * 30) *100   = 34.06 W

This Physics question is directed towards High School students and pertains to the maximum speed and height a 2-kg block achieves when released from a compressed spring, following the removal of a 3-kg block initially resting on it.

The subject of this question is Physics, and it is appropriate for a High School grade level. The question is regarding the oscillatory motion and energy conversion of a block attached to a spring system. Specifically, it involves understanding the concepts of potential energy stored in a spring, conservation of mechanical energy, and kinematics of simple harmonic motion.

To solve part (a) of the problem for the 2-kg block, you'll need to use the conservation of energy principle. Initially, when the 3-kg block is also on the spring, the potential energy stored in the compressed spring is equal to the kinetic energy the 2-kg block will have when it reaches its maximum speed. After the upper block is removed, the spring force will only accelerate the 2-kg block.

To find the maximum speed reached by the 2-kg block, you would calculate the kinetic energy equivalent to the potential energy stored in the spring and set it equal to \\((1/2)mv^2\\). For part (b), you could use the conservation of energy principle again to find the maximum height the 2-kg block reaches by equating the initial kinetic energy to the potential energy at the maximum height (\\(mgh\\)).

Answer:

Isothermal  expansion W₁ =-37198.9 J

Polytropic Compression W₂ =-34872.82 J

Isobaric Compression W₃ =  -6974.566 J

The net work for the cycle = -79046.29 J

Explanation:

Mass of air = 0.15 kg = 150 g

Molar mass = 28.9647 g/mol

Number of moles = 150 g /28.9647 g/mol = 5.179 moles of air

PV = nRT therefrore V = nRT/(P) = 5.179*8.314*(350+273.15)/(2×10⁶) = 0.0134167 m³

For isothermal expansion we have

P₁V₁ = P₂V₂ or V₂ = P₁V₁/P₂ = 2×10⁶*0.0134167 / (5×10⁵) = 0.0536668 m³

Therefore work done

W₁ = -nRTln(V₂/V₁) = -26833ln(4) = -37198.9 J

Stage 2

Compression polytropically we have

[tex]\frac{P_2}{P_3} = (\frac{V_3}{V_2} )^n[/tex]  where P₃ = 2 MPa

Therefore V₃ = [tex](\frac{1}{4} )^{\frac{1}{1.2} }*V_2[/tex]  = 1.6904×10⁻² m³

Work = W₂ = [tex]\frac{P_2V_2-P_3V_3}{n-1}[/tex] =  -34872.82 J

[tex]\frac{P_2}{P_3} = (\frac{T_2}{T_3} )^\frac{n}{n-1}[/tex]     or T₃ = [tex]T_2*(\frac{P_3}{P_2})^\frac{n-1}{n}[/tex] = 785.12 K

Isobaric compression we have  thus

Work done W₃ = P(V₁ -V₃) = -6974.566 J

Total work = W₁ + W₂ + W₃ = -37198.9 J + -34872.82 J + -6974.566 J = -79046.29 J