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2000 calories of heat are given to a thermodynamic system and the system does 3350 joule of external work. In this process the internal energy of the system is increased by 5030 joule . Calculate the value of the conversion factor J ?
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Introduction:
In this problem, we are given the amount of heat given to a thermodynamic system (2000 calories) and the amount of work done by the system (3350 Joules). We are also given that the internal energy of the system increased by 5030 Joules. We need to find the value of the conversion factor J.
Understanding the problem:
- Heat (Q) is the energy transferred to or from a system due to temperature difference.
- Work (W) is the energy transferred to or from a system due to a force acting through a distance.
- Internal energy (ΔU) is the total energy of a system and is the sum of the heat and work done on the system.
Solution:
1. We are given the amount of heat in calories, but we need to convert it to Joules, as the other values are given in Joules. 1 calorie is equal to 4.184 Joules.
- 2000 calories = 2000 * 4.184 Joules = 8368 Joules.
2. We can calculate the change in internal energy (ΔU) using the equation:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat transferred to the system, and W is the work done by the system.
- ΔU = 5030 Joules
- Q = 8368 Joules
- W = 3350 Joules
3. Substitute the given values into the equation:
5030 = 8368 - 3350
4. Simplify the equation:
5030 = 5018
5. This equation is not possible, which means there is an error in the given information or calculations.
Conclusion:
The given information or calculations must have an error because the equation does not hold true. Check the given values and calculations again to find the correct answer.
In this problem, we are given the amount of heat given to a thermodynamic system (2000 calories) and the amount of work done by the system (3350 Joules). We are also given that the internal energy of the system increased by 5030 Joules. We need to find the value of the conversion factor J.
Understanding the problem:
- Heat (Q) is the energy transferred to or from a system due to temperature difference.
- Work (W) is the energy transferred to or from a system due to a force acting through a distance.
- Internal energy (ΔU) is the total energy of a system and is the sum of the heat and work done on the system.
Solution:
1. We are given the amount of heat in calories, but we need to convert it to Joules, as the other values are given in Joules. 1 calorie is equal to 4.184 Joules.
- 2000 calories = 2000 * 4.184 Joules = 8368 Joules.
2. We can calculate the change in internal energy (ΔU) using the equation:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat transferred to the system, and W is the work done by the system.
- ΔU = 5030 Joules
- Q = 8368 Joules
- W = 3350 Joules
3. Substitute the given values into the equation:
5030 = 8368 - 3350
4. Simplify the equation:
5030 = 5018
5. This equation is not possible, which means there is an error in the given information or calculations.
Conclusion:
The given information or calculations must have an error because the equation does not hold true. Check the given values and calculations again to find the correct answer.
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2000 calories of heat are given to a thermodynamic system and the system does 3350 joule of external work. In this process the internal energy of the system is increased by 5030 joule . Calculate the value of the conversion factor J ?
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2000 calories of heat are given to a thermodynamic system and the system does 3350 joule of external work. In this process the internal energy of the system is increased by 5030 joule . Calculate the value of the conversion factor J ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 2000 calories of heat are given to a thermodynamic system and the system does 3350 joule of external work. In this process the internal energy of the system is increased by 5030 joule . Calculate the value of the conversion factor J ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2000 calories of heat are given to a thermodynamic system and the system does 3350 joule of external work. In this process the internal energy of the system is increased by 5030 joule . Calculate the value of the conversion factor J ?.
2000 calories of heat are given to a thermodynamic system and the system does 3350 joule of external work. In this process the internal energy of the system is increased by 5030 joule . Calculate the value of the conversion factor J ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 2000 calories of heat are given to a thermodynamic system and the system does 3350 joule of external work. In this process the internal energy of the system is increased by 5030 joule . Calculate the value of the conversion factor J ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2000 calories of heat are given to a thermodynamic system and the system does 3350 joule of external work. In this process the internal energy of the system is increased by 5030 joule . Calculate the value of the conversion factor J ?.
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