What happens to the temperature of the system(bomb calorimeter?) as the fuel releases energy? What principle does it follow? Discuss.

1. What happens to the temperature of the system(bomb calorimeter?) as the fuel releases energy? What principle does it follow? Discuss.

Bomb calorimeters have to withstand the large pressure within the calorimeter as the reaction is being measured. Electrical energy is used to ignite the fuel; as the fuel is burning, it will heat up the surrounding air, which expands and escapes through a tube that leads the air out of the calorimeter.

2. Why must the bomb calorimeter be submerged in water?

Answer:

(to saturate the internal atmosphere, thus ensuring that all water produced is liquid, and removing the need to include enthalpy of vaporization in calculations)

3. Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a mass of 1.30 kg and a specific heat of 3.41 J/(g°C). If the initial temperature of the calorimeter is 25.5°C, what is its final temperature?​

Answer:

30.9°

Explanation:

Data:

Heat = 24 KJ = 24,000 J

mass = 1.3kg = 1300 g

Cs = 3.41 J / (g °C)

Ti = 25.5 °C

Question: Tf

Formula: Heat = mCsΔT =>

ΔT = Heat / (mCs) = 24000J/[1300g*3.41J/(g°C)] = 5.4 °C

ΔT = Tf - Ti => Tf = ΔT + Ti = 25.5°C + 5.4°C = 30.9°C

Answer: 30.9°

4. Why is bomb calorimeter used in enthalpy change of combustion rather than coffe cup?

Answer:

A bomb calorimeter is used in the determination of the enthalpy change of combustion rather than a coffee cup calorimeter because it is designed specifically to measure the heat of combustion reactions, which are extremely exothermic and release a large amount of heat energy.

A bomb calorimeter is a sealed container that is filled with a known volume of oxygen, and it is equipped with a device that can ignite the sample inside under controlled conditions. The heat generated by the combustion reaction is transferred to the surrounding water, which increases in temperature. The rise in temperature of the water can be measured to calculate the heat of combustion of the substance.

On the other hand, a coffee cup calorimeter is a simple device that is used to measure the heat transferred between two bodies in contact, and it is not designed to measure the heat of combustion reactions, which are much more exothermic than most reactions that can be performed in a coffee cup calorimeter.

In summary, the bomb calorimeter is designed specifically to measure the heat of combustion of a substance, which is much more exothermic than most reactions that can be performed in a coffee cup calorimeter, and it also allows for a greater degree of control over the conditions under which the reaction takes place.

5. Explain the process of measuring the calorific value of a gaseous fuel using a bomb calorimeter. Discuss the role of each component of the calorimeter and the safety precautions that should be taken during the experiment.

Answer:

(answered by Knowledge to help and guide the questioners throughout)

A bomb calorimeter is an instrument that measures the calorific value (heat energy) of a gaseous fuel. It consists of a sealed metal chamber, known as a bomb, in which a sample of the fuel can be burned under controlled conditions. The heat released by the combustion reaction is collected by a body of water sitting inside the calorimeter, and the increase in temperature of the water is measured and used to calculate the energy released by the fuel.

The bomb calorimeter is made up of four main parts: the bomb, the oxygen inlet pump, the water container, and the thermometer. The bomb is an insulated steel container that can withstand high pressures, and is used to securely contain the fuel and control the combustion reaction. The oxygen inlet pump is used to create an oxygen-rich environment inside the bomb, which is necessary for complete combustion of the fuel. The water container is used to collect the heat released by the combustion reaction, and the thermometer is used to measure the increase in temperature of the water.

When measuring the calorific value of a gaseous fuel using a bomb calorimeter, there are several important safety precautions that must be taken. First, make sure the bomb is securely closed before starting the experiment. The contents of the bomb are highly flammable, and must not be exposed to open flames or sparks. Second, never leave the bomb unattended when it is in use, as it may overpressure and become dangerous. Third, use proper ventilation when the bomb is in use and keep combustible materials away from the calorimeter. Finally, always handle the calorimeter equipment carefully to avoid causing any unnecessary damage.

By following these safety guidelines and properly operating the bomb calorimeter, it is possible to accurately measure the calorific value of a gaseous fuel. In doing so, researchers can get valuable insight into the potential energy of the fuel and its effectiveness as a fuel source.

6. A 1.000 g sample of octane (C8H18) is burned in a bomb calorimeter containing 1200 grams of water at an initial temperature of 25.00°C. After the reaction, the final temperature of the water is 33.20°C. The heat capacity of the calorimeter (also known as the "calorimeter constant") is 837 J/°C. The specific heat of water is 4,184 J/g °C. Calculate the heat of combustion of octane in kJ/mol.​

Explanation:

: The heat of combustion of octane is, -5475.42KJ/mole−5475.42KJ/mole

Solution :

First we have to calculate the heat absorbed by water.

Formula used :

q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})q

w

=m

w

×c

w

×ΔT=m

w

×c

w

×(T

final

−T

initial

)

where,

q_wq

w

= heat absorbed = ?

m_wm

w

= mass of water = 1200 g

c_wc

w

= specific heat of water= 4.184J/g^oC4.184J/g

o

C

T_{final}T

final

= final temperature = 33.2^oC33.2

o

C

T_{initial}T

initial

= initial temperature = 25^oC25

o

C

Now put all the given values in the above formula, we get

q_w=1200g\times 4.184J/g^oC\times (33.2-25)^oCq

w

=1200g×4.184J/g

o

C×(33.2−25)

o

C

q_w=41170.56Jq

w

=41170.56J

Now we have to calculate the heat of calorimeter.

q_b=c_b\times (T_{final}-T_{initial})=837J/^oC\times (33.2-25)^oC=6863.4Jq

b

=c

b

×(T

final

−T

initial

)=837J/

o

C×(33.2−25)

o

C=6863.4J

Now we have to calculate the heat released in combustion.

\begin{gathered}q_{comb}=-(q_w+q_b)\\\\q_{comb}=-(41170.56+6863.4)\\\\q_{comb}=-48033.96J=-48.03KJ\end{gathered}

q

comb

=−(q

w

+q

b

)

q

comb

=−(41170.56+6863.4)

q

comb

=−48033.96J=−48.03KJ

Now we have to calculate the heat of combustion of octane.

As, 1 gram of octane released heat of combustion -48.03 KJ

So, 114 g/mole of octane released heat of combustion -48.03\times 114=-5475.42KJ/mole−48.03×114=−5475.42KJ/mole

Therefore, the heat of combustion of octane is, -5475.42KJ/mole−5475.42KJ/mole

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7. A 1.50-g sample of quinone (C6H4O2) is burned in a bomb calorimeter whose total heat capacity is 8.500 kJ/°C. The temperature of the calorimeter increases from 25.00 to 29.49°C. (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of quinone and per mole of quinone?​

Answer:

ano po?

Explanation:

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8. examples of isolated system

Answer:

For example, two balls rolling across a smooth surface to strike each other can be considered an isolated system, while two balls rolling across a gravel path or thick carpet are influenced by friction originating from outside the system and therefore are not an isolated system

Answer:

A thermos flask is the best example of an isolated system.

Explanation:

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9. A 2.200-g sample of quinone (c6h4o2) is burned in a bomb calorimeter whose total heat capacity is 7.857 kj/∘c. the temperature of the calorimeter increases from 23.44∘c to 30.57∘c. what is the heat of combustion per gram of quinone?

Answer:

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10. 4.4 g compound is burned in a bomb calorimeter whose total heat capacity is 7.854 kJ/0C. A temperature change of 7.130C occurred. What is the heat of combustion per gram of the compound?

Answer:

12.73 kJ/g

Explanation:

q=mCpdT

q/m= ((7.854kJ/degC) * 7.13 degC )/4.4g

q/m = 12.73 kJ/g

11. The isolated system is made up of the

Answer:

a physical system that is so isolated from other systems that it has no interaction with them. a thermodynamic system surrounded by immovable stiff barriers through which no mass nor energy can pass

12. The molar heat of combustion of a compound is 1,350 kJ/mol. If 0.875 moles of this compound was burned in a bomb calorimeter containing 1.70 L of water, what would the increase in temperature be?

Answer:

A heat capacity problem.

Know that,

- Specific heat of water = 4.184 kJ/kg•°C

- when burning the naphthalene = (0.821 g)(40.1 kJ/g) = 32.92 kJ  

- temperature rise is 4.21°C  

- temperature in heating 1000 g of water from (4.21°C )( 4.184 kJ/kg•°C) = 17.61kJ  

- heat energy required in heating the calorimeter (32.92 kJ)-(17.61 kJ) = 15.31 kJ  

- Lastly, divide heat energy from heated calorimeter to the rise of temperature = 15.31/4.21  

- heat capacity of calorimeter = 3.635 kJ/°C

Explanation:

13. it is the interaction that occurs in an isolated system?​

Answer:

An isolated system is a thermodynamic system that cannot exchange either energy or matter outside the boundaries of the system.

Explanation:

A system in which the only forces that contribute to the momentum change of an individual object are the forces acting between the objects themselves can be considered an isolated system.

14. The initial temperature of a bomb calorimeter is 28.50°C. When a chemist carries out a reaction in this calorimeter, its temperature decreases to 27.45°C. If the calorimeter has a mass of 1.400 kg and a specific heat of 3.52 J/(g°C), how much heat is absorbed by the reaction?

Answer:

Q = 5.17 × 10³ J

Explanation:

Calculating the heat absorbed

Q = mc∆T

Q = (1.400 × 10³ g)(3.52 J/(g°C))(28.50°C - 27.45°C)

Q = 5.17 × 10³ J

#CarryOnLearning

15. A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. The total heat capacity of the calorimeter plus the water was 5.760J ∘ C If the temperature rise of the calorimeter with water was 0.570 ∘ C calculate the enthalpy of combustion of magnesium. M g ( s ) + 1 2 O 2 ( g ) → M g O ( s )

Answer:

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16. To a sample of water at 25.8°C in a constant-pressure calorimeter of negligible heat capacity is added with a 14.2-g piece of aluminum whose temperature is 92.6 °C. If the final temperature of water is 27.3°C, calculate the mass of the water in the calorimeter. *Treat the calorimeter as an isolated system, which means no heat lost to the surroundings. * [ specific heat (s) of H2O = 4.184 J/g ⋅ C ] * [ specific heat (s) of Al = 0.900 J/g ⋅ C ]

The mass of the water in the calorimeter : m = 132.972 g

Further explanation  

The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released  

Q in = Q out  

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Water absorbs the heat from aluminum, so

Q water = Q aluminum

m x 4.184 J/g ⋅ C x (27.3 - 25.8) = 14.2 g x 0.9  J/g ⋅ C x (92.6 - 27.3)

m x 6.276 = 834.534

m = 132.972 g

Learn more  

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https://brainly.ph/question/1838433  

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#LetsStudy  

17. A 23.5 g sample of unknown metal at 79.5 °C was placed in a constant-pressure bomb calorimeter containing 50.0 g water at 30.0 °C. The temperature rises to 32.6 °C. What is the specific heat of the metal?​

The specific heat of the metal : 0.493 J/g° C  

Further explanation  

The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released  

Q in = Q out  

Qabsorbed+Qreleased=0

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Q metal + Q water = 0

metal⇒released heat

water⇒absorbed heat

[tex]\tt m.c.(t_f-t_i)metal+m.c.(t_f-t_i)water=0\\\\23.5.c(32.6-79.5)+50.4.184(32.6-30)=0\\\\-1102.15.c+543.92=0\\\\c=\dfrac{543.92}{1102.15}=\boxed{\bold{0.493~J/g^oC}}[/tex]

Learn more  

the heat capacity of the bomb calorimeter  

https://brainly.ph/question/1838433  

the combustion of sugar  

https://brainly.ph/question/1968331  

the heat capacity of a liquid and solid  

https://brainly.ph/question/1652238  

#LetsStudy  

18. in calorimeter experiment, discuss the parts of your set-up. what are included as the system and which are the surroundings?

In calorimeter experiment, discuss the parts of your set-up. What are included as the system and which are the surroundings?

In a calorimeter experiment, the set-up includes several parts such as the calorimeter, the sample to be tested, and the thermometer. The system refers to the part of the experiment being studied, which is the sample in this case. The surroundings refer to everything else in the experiment, including the calorimeter and the thermometer.

The calorimeter is a device that is used to measure the heat absorbed or released during a chemical reaction or physical change. It is usually made of two nested containers, with the space between them filled with an insulating material to minimize heat transfer with the surroundings. The inner container holds the sample to be tested, and the outer container acts as a heat shield to protect the sample from temperature changes in the surroundings.

The sample being tested is considered the system in the experiment. It is the part of the experiment that is being studied to determine the amount of heat absorbed or released. This can be a solid, liquid, or gas, depending on the nature of the experiment.

The thermometer is used to measure the temperature changes in the sample and the surroundings. It is placed in the inner container along with the sample to monitor the temperature changes during the experiment.

In summary, the parts of a calorimeter experiment include the calorimeter, the sample being tested (system), and the thermometer. The surroundings refer to everything else in the experiment that is not part of the system, including the calorimeter and the thermometer.

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19. When 1.0 g of fructose, C6H12O6(s), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion?

Answer:

idol pwede ka ba isale sa raffle namen pwede ka manalo Ng 500 lods free idol

20. Write the balance chemical equation for the bomb calorimeter reaction of quinone (C6H4O2)

Answer:

pano po iyan?

Explanation:

com chem ba ito?

21. Acetylene, C2H2, is used in welding torches. It releases a lot of energy when burned in oxygen. The combustion of one gram of acetylene releases 48. 2 kJ. A 0. 750-gram sample of acetylene is burned in a bomb calorimeter (heat capacity = 1. 117kJ/°C) that contains 800. 0 g of water. The final temperature of the bomb and water after combustion is 35. 2°C. What is the initial temperature of the bomb and water?.

35.2°c is the initial temp of the bomb and water

22. Naphthalene combustion can be used to calibrate the heat capacity of a bomb calorimeter. the heat of combustion of naphthalene is –40.1 kj/g. when 0.8210 g of naphthalene was burned in a calorimeter containing 1,000. g of water, a temperature rise of 4.21°c was observed. what is the heat capacity of the bomb calorimeter excluding the water?

A heat capacity problem.

Know that,

- Specific heat of water = 4.184 kJ/kg•°C

- when burning the naphthalene = (0.821 g)(40.1 kJ/g) = 32.92 kJ  

- temperature rise is 4.21°C  

- temperature in heating 1000 g of water from (4.21°C )( 4.184 kJ/kg•°C) = 17.61kJ  

- heat energy required in heating the calorimeter (32.92 kJ)-(17.61 kJ) = 15.31 kJ  

- Lastly, divide heat energy from heated calorimeter to the rise of temperature = 15.31/4.21  

- heat capacity of calorimeter = 3.635 kJ/°C

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23. Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a mass of 1.30 kg and a specific heat of 3.41 J/(g°C). If the initial temperature of the calorimeter is 25.5°C, what is its final temperature?

Answer:

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24. In the bomb calorimetry simulation why is the energy released by the iron wire included in the calorimeter constant but deducted in the calculation of the heat of combustion of the substance?​

Answer:

In a bomb calorimetry simulation, the energy released by the iron wire is included in the calorimeter constant because it represents the heat absorbed or released by the calorimeter itself. The calorimeter constant is used to account for any heat loss or gain that occurs within the calorimeter itself, and it is used to correct the measured heat of combustion of the substance being tested.

The energy released by the iron wire is deducted in the calculation of the heat of combustion of the substance because it is not directly related to the combustion of the substance. The heat of combustion is a measure of the energy released by the complete combustion of a substance, and it should only include the energy released by the substance itself. The energy released by the iron wire is not related to the combustion of the substance and therefore should not be included in the heat of combustion calculation.

In summary, the energy released by the iron wire is included in the calorimeter constant because it represents the heat absorbed or released by the calorimeter, but it is deducted in the calculation of the heat of combustion of the substance because it is not directly related to the combustion of the substance.

Explanation:

25. he combustion of toluene has a ΔErxn of -3.91 x 10³ kJ/mol. When 1.55 g of toluene (C₇H₈) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. Find the heat capacity of the bomb calorimeter.

Answer:

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26. When 1.0 g of fructose, C6H12O6(s), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion?

Answer:

Correct option is

A

−2808kJ/mol

Heat released by the combustion =C

p

×ΔT=10×1.56 kJ=15.6 kJ

Hence, ΔH=−15.6 kJ/g

Molar mass of fructose =180g

Hence, ΔH

combustion

=−15.6×180 kJ/mol=−2808 kJ/mol

27. During the melting of the ice cube in the calorimeter, identify which is the system, the surroundings, and the boundary.

Answer:

As ice melts into water, kinetic energy is being added to the particles. This causes them to be 'excited' and they break the bonds that hold them together as a solid, resulting in a change of state: solid -> liquid.

28. )The initial temperature of a bomb calorimeter is 28.50°C. When a chemist carries out a reaction in this calorimeter, its temperature decreases to 27.45°C. If the calorimeter has a mass of 1.400 kg and a specific heat of 3.52 J/(g°C), how much heat is absorbed by the reaction?

Answer:

12.00 to 13.50c

Explanation:

#carry on learning

29. A 12.7 g sample of sulfur (S8) is placed in a bomb that is then filled with oxygen under pressure. The bomb is placed in the calorimeter that is filled with 2.20 kg of water at 21.08°C. The reaction mixture is ignited and the temperature rises to a high of 33.88°C. From this data, calculate the molar heat of combustion of sulfur.

Assumptions:

Negligible heat dissipated to the container

[tex]Q=mCp \Delta T\\Q=(2.20 \ kg)(4.184 \ \frac{J}{g-^0C/K})(33.88-21.08)\\Q=+117.8214 \ J\\[/tex]

This is the energy absorbed by the water. Thus the energy absorbed by the water is equal to the energy released by the reaction (with certain assumptions).

[tex]\frac{117.8214 \ J}{12.7 \ g \ sulfur}\times (\frac{64.0638 \ g}{mol \ sulfur})=594.3375 \ \frac{J}{mol}[/tex]

S-e=A That a sulfur for a %*H and B

30. 1. To begin the experiment, 1.11g of methane CH 4 is burned in a bomb calorimeter containing 1000 grams of water. The initial temperature of water is 24.85 °C. The specific heat of water is 4.184 J/g °C. The heat capacity of the calorimeter is 695 J/°C. After the reaction the final temperature of the water is 35.65 °C. calculate the total heat combustion

Answer:

26.19 grams of water can help calculate the temperature