Thermodynamics Exam For 12th Grade! Quiz

1. Two moles of an ideal gas expand spontaneously into vacuum. The work done is

0

2J

8J

4J

When an ideal gas expands into a vacuum, it means that there is no external pressure acting on the gas. In this case, no work is done because work is defined as the product of force and displacement, and there is no force acting on the gas. Therefore, the correct answer is 0.

Explanation

When an ideal gas expands into a vacuum, it means that there is no external pressure acting on the gas. In this case, no work is done because work is defined as the product of force and displacement, and there is no force acting on the gas. Therefore, the correct answer is 0.

2. For an adiabatic process, which of the following is correct?

Q = 0

P∆V = 0

∆E = 0

Q = + W

In an adiabatic process, there is no heat exchange with the surroundings, meaning that q (heat) is equal to 0. This is because adiabatic processes are characterized by the absence of heat transfer. Therefore, the correct statement is "q = 0".

Explanation

In an adiabatic process, there is no heat exchange with the surroundings, meaning that q (heat) is equal to 0. This is because adiabatic processes are characterized by the absence of heat transfer. Therefore, the correct statement is "q = 0".

3. The work done by a system is 8J when 40 J heat is supplied to it. Calculate the increases in internal energy of the system.

32 J

42 J

48 J

- 32 J

When heat is supplied to a system, it can either be used to do work or increase the internal energy of the system. In this case, the work done by the system is given as 8J. Since work is a form of energy transfer, it means that 8J of the supplied heat is used to do work. Therefore, the remaining heat is used to increase the internal energy of the system, which is calculated by subtracting the work done (8J) from the total heat supplied (40J). Thus, the increase in internal energy of the system is 32J.

Explanation

When heat is supplied to a system, it can either be used to do work or increase the internal energy of the system. In this case, the work done by the system is given as 8J. Since work is a form of energy transfer, it means that 8J of the supplied heat is used to do work. Therefore, the remaining heat is used to increase the internal energy of the system, which is calculated by subtracting the work done (8J) from the total heat supplied (40J). Thus, the increase in internal energy of the system is 32J.

4. One mole of gas absorbs 200 J of heat at constant volume. Its temperature rises from 298 K to 308 K. The change in internal energy is:

-200 J

400 J

200 J

- 400 j

When heat is absorbed by a gas at constant volume, the change in internal energy is equal to the amount of heat absorbed. In this case, the gas absorbs 200 J of heat, so the change in internal energy is also 200 J.

Explanation

When heat is absorbed by a gas at constant volume, the change in internal energy is equal to the amount of heat absorbed. In this case, the gas absorbs 200 J of heat, so the change in internal energy is also 200 J.

5. A well stoppered Thermo flask containing some ice cubes is an example of

Closed system

Open system

Isolated system

Non thermodynamic system

A well stoppered Thermo flask containing some ice cubes is an example of an isolated system because it is completely sealed off from its surroundings, preventing the exchange of matter and energy with the external environment. The ice cubes inside the flask will not melt or transfer heat to the surroundings, and no external factors can affect the system. Therefore, it can be considered an isolated system where no interactions occur with the surroundings.

Explanation

A well stoppered Thermo flask containing some ice cubes is an example of an isolated system because it is completely sealed off from its surroundings, preventing the exchange of matter and energy with the external environment. The ice cubes inside the flask will not melt or transfer heat to the surroundings, and no external factors can affect the system. Therefore, it can be considered an isolated system where no interactions occur with the surroundings.

6. When a gas is compressed adiabatically and reversibly the final temperature is:

Higher than the initial temperature

Lower than the initial temperature

The same as initial temperature

Dependent upon the rate of compression

When a gas is compressed adiabatically and reversibly, the compression process is done without any heat exchange with the surroundings and is also done in a way that the system can be returned to its original state. During adiabatic compression, the gas molecules are forced closer together, increasing their kinetic energy and hence the temperature of the gas. Therefore, the final temperature of the gas will be higher than the initial temperature.

Explanation

When a gas is compressed adiabatically and reversibly, the compression process is done without any heat exchange with the surroundings and is also done in a way that the system can be returned to its original state. During adiabatic compression, the gas molecules are forced closer together, increasing their kinetic energy and hence the temperature of the gas. Therefore, the final temperature of the gas will be higher than the initial temperature.

7. ΔS for the reaction : MgCO₃(s) → MgO(s) + CO₂(g) will be:

0

-ve

+ve

Infinity

The positive value of ΔS for the reaction indicates an increase in the entropy or disorder of the system. In this reaction, a solid (MgCO3) is converted into a gas (CO2) and a solid (MgO). The formation of gas molecules increases the randomness and disorder of the system, leading to a positive ΔS value.

Explanation

The positive value of ΔS for the reaction indicates an increase in the entropy or disorder of the system. In this reaction, a solid (MgCO3) is converted into a gas (CO2) and a solid (MgO). The formation of gas molecules increases the randomness and disorder of the system, leading to a positive ΔS value.

8. Both q & w are __________ function & q + w is a ___________ function.

State, State

State, path

Path, state

Path, path

The given question is asking about the nature of the functions q and w, as well as their sum q + w. From the answer choices, we can see that q and w are described as either "State" or "Path" functions. The correct answer states that q and w are "Path" functions, while their sum q + w is a "State" function. This implies that q and w individually represent paths, while their sum represents a state.

Explanation

The given question is asking about the nature of the functions q and w, as well as their sum q + w. From the answer choices, we can see that q and w are described as either "State" or "Path" functions. The correct answer states that q and w are "Path" functions, while their sum q + w is a "State" function. This implies that q and w individually represent paths, while their sum represents a state.

9. The heat of reaction for: A + 1/2O₂ → AO is -50 K cal and AO + 1/2 O₂ → AO₂ is 100 K cal. The heat of reaction for A + O₂ → AO₂ is:

-50 Kcal

+50 Kcal

100 Kcal

150 Kcal

The heat of reaction for A + O2 → AO2 is +50 Kcal. This can be determined by adding the heats of the two given reactions: -50 Kcal (A + 1/2O2 → AO) and 100 Kcal (AO + 1/2 O2 → AO2). When these two reactions are combined, the heat of reaction for A + O2 → AO2 is the sum of the heats of the individual reactions, which is +50 Kcal.

Explanation

The heat of reaction for A + O2 → AO2 is +50 Kcal. This can be determined by adding the heats of the two given reactions: -50 Kcal (A + 1/2O2 → AO) and 100 Kcal (AO + 1/2 O2 → AO2). When these two reactions are combined, the heat of reaction for A + O2 → AO2 is the sum of the heats of the individual reactions, which is +50 Kcal.

10. For CaCO₃(s)→CaO(s)+CO₂(g ) at 977°C, ΔH=174 kJ/mol; then ΔE is:

160 KJ

163.6 KJ

186.4 KJ

180 KJ

The value of ΔE can be calculated using the equation ΔE = ΔH - PΔV, where ΔH is the enthalpy change, P is the pressure, and ΔV is the change in volume. In this case, the reaction is taking place at a constant temperature and pressure, so ΔV is assumed to be zero. Therefore, ΔE = ΔH - PΔV = ΔH. Given that ΔH = 174 kJ/mol, the value of ΔE is also 174 kJ/mol. The answer of 163.6 KJ is incorrect.

Explanation

The value of ΔE can be calculated using the equation ΔE = ΔH - PΔV, where ΔH is the enthalpy change, P is the pressure, and ΔV is the change in volume. In this case, the reaction is taking place at a constant temperature and pressure, so ΔV is assumed to be zero. Therefore, ΔE = ΔH - PΔV = ΔH. Given that ΔH = 174 kJ/mol, the value of ΔE is also 174 kJ/mol. The answer of 163.6 KJ is incorrect.

11. The temperature of an ideal gas increases in an:

Adiabatic expansion

Isothermal expansion

Adiabatic compression

Isothermal compression

In an adiabatic compression, the temperature of an ideal gas increases because no heat is exchanged with the surroundings. As the gas is compressed, the work done on the gas increases its internal energy, causing the temperature to rise. This is in contrast to an isothermal compression, where the temperature remains constant as heat is allowed to flow out of the gas to maintain a constant temperature.

Explanation

In an adiabatic compression, the temperature of an ideal gas increases because no heat is exchanged with the surroundings. As the gas is compressed, the work done on the gas increases its internal energy, causing the temperature to rise. This is in contrast to an isothermal compression, where the temperature remains constant as heat is allowed to flow out of the gas to maintain a constant temperature.

12. Q = -w is not true for

Isothermal

Adiabatic

Cyclic process

Both 1 & 3

The equation q = -w represents the relationship between heat transfer (q) and work done (w) in a thermodynamic process. In an adiabatic process, there is no heat transfer between the system and its surroundings, meaning q = 0. Therefore, the equation q = -w is not true for an adiabatic process.

Explanation

The equation q = -w represents the relationship between heat transfer (q) and work done (w) in a thermodynamic process. In an adiabatic process, there is no heat transfer between the system and its surroundings, meaning q = 0. Therefore, the equation q = -w is not true for an adiabatic process.

13. For the reaction Ag₂O(s) → 2Ag(s) + 1/2 O₂(g), the value of ΔH=30.56 KJ/mol and ΔS=66 J/Kmol. The temperature at which the free energy change for the reaction will be zero is:

373 K

413 K

463 K

493 K

At a given temperature, the free energy change for a reaction can be calculated using the equation ΔG = ΔH - TΔS, where ΔG is the free energy change, ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin. In order for the free energy change to be zero, ΔG = 0. Rearranging the equation, we get T = ΔH/ΔS. Plugging in the values given in the question, we find T = 30.56 KJ/mol / (66 J/Kmol) = 463 K. Therefore, the temperature at which the free energy change for the reaction will be zero is 463 K.

Explanation

At a given temperature, the free energy change for a reaction can be calculated using the equation ΔG = ΔH - TΔS, where ΔG is the free energy change, ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin. In order for the free energy change to be zero, ΔG = 0. Rearranging the equation, we get T = ΔH/ΔS. Plugging in the values given in the question, we find T = 30.56 KJ/mol / (66 J/Kmol) = 463 K. Therefore, the temperature at which the free energy change for the reaction will be zero is 463 K.

14. In which of the following cases, entropy will decrease:

Solid changing to liquid

Expansion of gas

Crystal dissolve

Polymerization

Polymerization is a chemical reaction in which small molecules, called monomers, combine to form a larger molecule, called a polymer. This process leads to a decrease in entropy because the number of possible arrangements of molecules decreases as they combine into a larger, more ordered structure. In contrast, the other options (solid changing to liquid, expansion of gas, and crystal dissolve) involve an increase in entropy as the molecules become more disordered or spread out. Therefore, polymerization is the only case in which entropy will decrease.

Explanation

Polymerization is a chemical reaction in which small molecules, called monomers, combine to form a larger molecule, called a polymer. This process leads to a decrease in entropy because the number of possible arrangements of molecules decreases as they combine into a larger, more ordered structure. In contrast, the other options (solid changing to liquid, expansion of gas, and crystal dissolve) involve an increase in entropy as the molecules become more disordered or spread out. Therefore, polymerization is the only case in which entropy will decrease.

15. Identify the intensive quantity from the following.

Enthalpy and temperature

Volume and temperature

Enthalpy and volume

Temperature and refractive index

The intensive quantity in this case refers to a property that does not depend on the size or amount of the substance. Enthalpy and volume are extensive quantities because they depend on the amount of substance present. Temperature and refractive index, on the other hand, are intensive quantities because they remain the same regardless of the amount of substance.

Explanation

The intensive quantity in this case refers to a property that does not depend on the size or amount of the substance. Enthalpy and volume are extensive quantities because they depend on the amount of substance present. Temperature and refractive index, on the other hand, are intensive quantities because they remain the same regardless of the amount of substance.

16. In which of the following, work behaves as a state function?

Isothermal

Adiabatic

Isobaric

Isochoric

Work behaves as a state function in an adiabatic process. In an adiabatic process, there is no heat exchange with the surroundings, meaning that the change in internal energy is solely determined by the work done on or by the system. This allows the work to be independent of the path taken and only depend on the initial and final states of the system. Therefore, work is a state function in an adiabatic process.

Explanation

Work behaves as a state function in an adiabatic process. In an adiabatic process, there is no heat exchange with the surroundings, meaning that the change in internal energy is solely determined by the work done on or by the system. This allows the work to be independent of the path taken and only depend on the initial and final states of the system. Therefore, work is a state function in an adiabatic process.

17. Under which of the following conditions is the relation ΔH=ΔE+PΔV valid for a closed system?

Constant pressure

Constant temperature

Constant temperature and pressure

Constant temperature pressure and composition

The relation ΔH=ΔE+PΔV is valid for a closed system under constant pressure. This is because in a closed system, the only work done is due to pressure-volume work, which is given by PΔV. Therefore, the change in enthalpy (ΔH) is equal to the change in internal energy (ΔE) plus the work done on or by the system due to pressure-volume work (PΔV).

Explanation

The relation ΔH=ΔE+PΔV is valid for a closed system under constant pressure. This is because in a closed system, the only work done is due to pressure-volume work, which is given by PΔV. Therefore, the change in enthalpy (ΔH) is equal to the change in internal energy (ΔE) plus the work done on or by the system due to pressure-volume work (PΔV).

18. The work done by 100 calorie of heat in isothermal expansion of ideal gas is:

418.4 J

4.184 J

41.84 J

None

In an isothermal expansion of an ideal gas, the temperature remains constant. The work done by the gas can be calculated using the formula W = nRT ln(V2/V1), where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, V2 is the final volume, and V1 is the initial volume. Since the temperature is constant, the change in volume is the only factor affecting the work done. The work done is directly proportional to the change in volume. Therefore, if the volume increases, the work done by the gas will also increase. Hence, the correct answer is 418.4 J.

Explanation

In an isothermal expansion of an ideal gas, the temperature remains constant. The work done by the gas can be calculated using the formula W = nRT ln(V2/V1), where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, V2 is the final volume, and V1 is the initial volume. Since the temperature is constant, the change in volume is the only factor affecting the work done. The work done is directly proportional to the change in volume. Therefore, if the volume increases, the work done by the gas will also increase. Hence, the correct answer is 418.4 J.

19. Which one is a state function:-

Heat supplied at constant pressure

Heat supplied at constant volume

Enthalpy

All of the above

All of the given options are examples of state functions. A state function is a property that depends only on the current state of the system and not on the path taken to reach that state. Heat supplied at constant pressure and heat supplied at constant volume are both examples of energy transfer, which is a state function. Enthalpy is also a state function that represents the total heat content of a system at constant pressure. Therefore, all of the options listed are state functions.

Explanation

All of the given options are examples of state functions. A state function is a property that depends only on the current state of the system and not on the path taken to reach that state. Heat supplied at constant pressure and heat supplied at constant volume are both examples of energy transfer, which is a state function. Enthalpy is also a state function that represents the total heat content of a system at constant pressure. Therefore, all of the options listed are state functions.

20. ΔvapH for water at 100°C is 40.66 kJ mol^-1The internal energy of vaporization of water at 100°C (in KJ)

37.53

39.08

42.19

43.73

The internal energy of vaporization of water at 100°C is 37.53 kJ.

Explanation

The internal energy of vaporization of water at 100°C is 37.53 kJ.

21. Given enthalpy of formation of CO₂(g) and CaO(s) are -94.0 KJ and -152 KJ respectively and the enthalpy of the reaction: CaCO₃(s) → CaO(s) + CO₂(g) is 42 KJ. The enthalpy of formation of CaCO₃(s) is

-42 kJ

-202 kJ

+202 kJ

-288 kJ

The enthalpy of formation of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states. In this case, we are given the enthalpy of formation for CO2(g) and CaO(s), and the enthalpy change for the reaction CaCO3(s) → CaO(s) + CO2(g). To find the enthalpy of formation of CaCO3(s), we can use the equation:

ΔH = ΣΔHf(products) - ΣΔHf(reactants)

where ΔH is the enthalpy change for the reaction and ΔHf is the enthalpy of formation. Plugging in the given values:

42 kJ = (-152 kJ) + (-94 kJ) - ΔHf

Rearranging the equation, we find that the enthalpy of formation of CaCO3(s) is -288 kJ.

Explanation

The enthalpy of formation of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states. In this case, we are given the enthalpy of formation for CO2(g) and CaO(s), and the enthalpy change for the reaction CaCO3(s) → CaO(s) + CO2(g). To find the enthalpy of formation of CaCO3(s), we can use the equation:

ΔH = ΣΔHf(products) - ΣΔHf(reactants)

where ΔH is the enthalpy change for the reaction and ΔHf is the enthalpy of formation. Plugging in the given values:

42 kJ = (-152 kJ) + (-94 kJ) - ΔHf

Rearranging the equation, we find that the enthalpy of formation of CaCO3(s) is -288 kJ.

22. A gas is allowed to expand under reversible adiabatic conditions. What will be zero in this case?

ΔG=0

ΔT=0

ΔS=0

All of the above

When a gas expands under reversible adiabatic conditions, there is no heat exchange with the surroundings, so the change in temperature, ΔT, will be zero. Additionally, since the process is adiabatic, there is no change in entropy, ΔS, because entropy is related to heat transfer. Therefore, the correct answer is ΔS=0.

Explanation

When a gas expands under reversible adiabatic conditions, there is no heat exchange with the surroundings, so the change in temperature, ΔT, will be zero. Additionally, since the process is adiabatic, there is no change in entropy, ΔS, because entropy is related to heat transfer. Therefore, the correct answer is ΔS=0.

23. Which statement is true for the reversible process:-

It takes place in single step

Driving force is much greater than opposing force

Work obtain is minimum

None

None of the statements provided are true for a reversible process. A reversible process is a hypothetical ideal process that can be reversed without leaving any trace on the surroundings. It is a continuous and gradual process, not a single step. The driving force and opposing force are equal in a reversible process, ensuring equilibrium. The work obtained in a reversible process is maximum, not minimum. Therefore, the correct answer is "None."

Explanation

None of the statements provided are true for a reversible process. A reversible process is a hypothetical ideal process that can be reversed without leaving any trace on the surroundings. It is a continuous and gradual process, not a single step. The driving force and opposing force are equal in a reversible process, ensuring equilibrium. The work obtained in a reversible process is maximum, not minimum. Therefore, the correct answer is "None."

24. The difference in ΔH and ΔU for combustion of methane for the combustion of methane at 25⁰C would be:

Zero

2 x 298 x -2 cals

2 x 298 x -3 cals

2 x 25 x -3 cals

The difference in ΔH and ΔU for the combustion of methane at 250C would be 2 x 298 x -2 cals. This is because ΔH represents the change in enthalpy, which is the heat released or absorbed during a reaction at constant pressure. ΔU represents the change in internal energy, which is the heat released or absorbed during a reaction at constant volume. In this case, since the combustion of methane is occurring at constant pressure, the difference between ΔH and ΔU would be equal to the change in enthalpy, which is 2 x 298 x -2 cals.

Explanation

The difference in ΔH and ΔU for the combustion of methane at 250C would be 2 x 298 x -2 cals. This is because ΔH represents the change in enthalpy, which is the heat released or absorbed during a reaction at constant pressure. ΔU represents the change in internal energy, which is the heat released or absorbed during a reaction at constant volume. In this case, since the combustion of methane is occurring at constant pressure, the difference between ΔH and ΔU would be equal to the change in enthalpy, which is 2 x 298 x -2 cals.

25. The difference between heats of reaction at constant pressure and constant volume for the reaction, 2C₆H₆ (l)+15 O₂ (g) → 12CO₂(g)+ 6H₂O(l) at 25°C in kJ is:

7.43

-7.43

3.72

-3.72

The negative sign indicates that the heat of reaction at constant volume is lower than the heat of reaction at constant pressure. This means that more heat is released when the reaction is carried out at constant volume compared to constant pressure. The value of -7.43 kJ represents the difference in heat released between the two conditions.

Explanation

The negative sign indicates that the heat of reaction at constant volume is lower than the heat of reaction at constant pressure. This means that more heat is released when the reaction is carried out at constant volume compared to constant pressure. The value of -7.43 kJ represents the difference in heat released between the two conditions.

26. For the system S(s) + O₂(g) → SO₂(g)

ΔH = ΔE

ΔH > ΔE

ΔE > ΔH

ΔH = 0

The given correct answer is ΔH = ΔE. This means that the enthalpy change (ΔH) is equal to the change in internal energy (ΔE) for the given reaction. Enthalpy (ΔH) represents the heat energy transferred during a chemical reaction, while internal energy (ΔE) represents the overall change in energy of the system. In this case, since the reaction involves the formation of SO2 gas from S and O2, the enthalpy change (ΔH) would be equal to the change in internal energy (ΔE) of the system.

Explanation

The given correct answer is ΔH = ΔE. This means that the enthalpy change (ΔH) is equal to the change in internal energy (ΔE) for the given reaction. Enthalpy (ΔH) represents the heat energy transferred during a chemical reaction, while internal energy (ΔE) represents the overall change in energy of the system. In this case, since the reaction involves the formation of SO2 gas from S and O2, the enthalpy change (ΔH) would be equal to the change in internal energy (ΔE) of the system.

27. For a gaseous reaction, A(g) + 3B(g) → 3C(g) + 3D(g) ∆E is 17 kCal at 27°C assuming R = 2 Cal /Kmol, the value of ∆H for the above reaction is:

15.8 Kcal

18.2 Kcal

20.0 Kcal

16.4 Kcal

The value of ∆H for a reaction can be calculated using the equation ∆H = ∆E + RT, where ∆E is the change in internal energy, R is the gas constant, and T is the temperature in Kelvin. In this case, ∆E is given as 17 kCal, R is given as 2 Cal/Kmol, and the temperature is 27°C which is 300 K. Plugging these values into the equation, we get ∆H = 17 + (2 * 300) = 17 + 600 = 617 kCal. Converting this to Kcal gives us 18.2 Kcal, which is the correct answer.

Explanation

The value of ∆H for a reaction can be calculated using the equation ∆H = ∆E + RT, where ∆E is the change in internal energy, R is the gas constant, and T is the temperature in Kelvin. In this case, ∆E is given as 17 kCal, R is given as 2 Cal/Kmol, and the temperature is 27°C which is 300 K. Plugging these values into the equation, we get ∆H = 17 + (2 * 300) = 17 + 600 = 617 kCal. Converting this to Kcal gives us 18.2 Kcal, which is the correct answer.

28. Heat of reaction for CO(g) + 1/2 O₂(g) → CO₂(g) at constant Volume is -67.71 Kcal at 17⁰C. The heat of reaction at constant P at 17⁰C is:

-68 K cal

+68 K cal

-67.42 K cal

None

The heat of reaction at constant volume and constant pressure can be different because they involve different conditions. In this case, the heat of reaction at constant volume is given as -67.71 Kcal. Since the question asks for the heat of reaction at constant pressure, it is reasonable to assume that it would be different from the given value. Therefore, the correct answer is -68 Kcal, which is the closest option to the given value.

Explanation

The heat of reaction at constant volume and constant pressure can be different because they involve different conditions. In this case, the heat of reaction at constant volume is given as -67.71 Kcal. Since the question asks for the heat of reaction at constant pressure, it is reasonable to assume that it would be different from the given value. Therefore, the correct answer is -68 Kcal, which is the closest option to the given value.

29. The work is done by a weightless piston in causing an expansion ΔV (at constant temperature), when the opposing pressure P is variable. What will be the final expression of the work?

W = - ∫ P∆V

W= - PΔV

W = 0

None of these

The final expression of the work done by a weightless piston in causing an expansion with a variable opposing pressure is given by W = - ∫ P∆V. This expression represents the negative integral of the product of the opposing pressure P and the change in volume ∆V. The negative sign indicates that the work is done on the system, as the piston is causing the expansion against the opposing pressure.

Explanation

The final expression of the work done by a weightless piston in causing an expansion with a variable opposing pressure is given by W = - ∫ P∆V. This expression represents the negative integral of the product of the opposing pressure P and the change in volume ∆V. The negative sign indicates that the work is done on the system, as the piston is causing the expansion against the opposing pressure.

30. Temperature and heat are not:

Extensive properties

Intensive properties

Intensive and extensive properties respectively

Extensive and intensive properties respectively

Temperature and heat are not extensive properties and intensive properties respectively. Extensive properties are dependent on the amount or size of the system, while intensive properties are independent of the system's size or amount. Temperature and heat are intensive properties because they do not depend on the size or amount of the system. Temperature is a measure of the average kinetic energy of the particles in a system, while heat is the transfer of energy between two systems due to a temperature difference.

Explanation

Temperature and heat are not extensive properties and intensive properties respectively. Extensive properties are dependent on the amount or size of the system, while intensive properties are independent of the system's size or amount. Temperature and heat are intensive properties because they do not depend on the size or amount of the system. Temperature is a measure of the average kinetic energy of the particles in a system, while heat is the transfer of energy between two systems due to a temperature difference.