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The contents of a well-insulated tank are heated by a resistor of 2Ω in which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat (Q), work (W) and change in internal energy (ΔU) during the process in kW are
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The contents of a well-insulated tank are heated by a resistor of 2&Om...
electric work = I2R = 2.3 kW
On the system ⇒ W = -2.3 kW
& ΔU = -W = 2.3 kW
Most Upvoted Answer
The contents of a well-insulated tank are heated by a resistor of 2&Om...
Understanding the Problem
In a well-insulated system, we analyze the effects of electric heating on the internal energy of the tank's contents. The resistor has a resistance of 23 ohms, and a current of 10 A is flowing.
Calculating Heat Transfer (Q)
- The power (P) provided by the resistor can be calculated using the formula:
\[
P = I^2 R
\]
Where:
- I = 10 A
- R = 23 ohms
- Therefore:
\[
P = (10^2) \times 23 = 100 \times 23 = 2300 \text{ W} = 2.3 \text{ kW}
\]
- In terms of heat transfer (Q), since the system is gaining energy through heating:
\[
Q = +2.3 \text{ kW}
\]
Work Done by the System (W)
- In this case, the system does not perform any work on the surroundings, as it is insulated. Therefore:
\[
W = 0 \text{ kW}
\]
Change in Internal Energy (ΔU)
- According to the first law of thermodynamics:
\[
\Delta U = Q - W
\]
- Substituting the values:
\[
\Delta U = +2.3 \text{ kW} - 0 \text{ kW} = +2.3 \text{ kW}
\]
Conclusion
- The final values are:
- **Heat Transfer (Q)**: +2.3 kW
- **Work Done (W)**: 0 kW
- **Change in Internal Energy (ΔU)**: +2.3 kW
Thus, the correct answer is option **b**: Q = +2.3, W = 0, ΔU = +2.3.
However, if option A was stated as correct, please verify the problem context; likely, it is a misunderstanding of the sign conventions used in the context.
In a well-insulated system, we analyze the effects of electric heating on the internal energy of the tank's contents. The resistor has a resistance of 23 ohms, and a current of 10 A is flowing.
Calculating Heat Transfer (Q)
- The power (P) provided by the resistor can be calculated using the formula:
\[
P = I^2 R
\]
Where:
- I = 10 A
- R = 23 ohms
- Therefore:
\[
P = (10^2) \times 23 = 100 \times 23 = 2300 \text{ W} = 2.3 \text{ kW}
\]
- In terms of heat transfer (Q), since the system is gaining energy through heating:
\[
Q = +2.3 \text{ kW}
\]
Work Done by the System (W)
- In this case, the system does not perform any work on the surroundings, as it is insulated. Therefore:
\[
W = 0 \text{ kW}
\]
Change in Internal Energy (ΔU)
- According to the first law of thermodynamics:
\[
\Delta U = Q - W
\]
- Substituting the values:
\[
\Delta U = +2.3 \text{ kW} - 0 \text{ kW} = +2.3 \text{ kW}
\]
Conclusion
- The final values are:
- **Heat Transfer (Q)**: +2.3 kW
- **Work Done (W)**: 0 kW
- **Change in Internal Energy (ΔU)**: +2.3 kW
Thus, the correct answer is option **b**: Q = +2.3, W = 0, ΔU = +2.3.
However, if option A was stated as correct, please verify the problem context; likely, it is a misunderstanding of the sign conventions used in the context.
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The contents of a well-insulated tank are heated by a resistor of 2Ωin which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive.The rates of heat (Q), work (W) and change in internal energy (ΔU)during the process in kW area)b)c)d)Correct answer is option 'A'. Can you explain this answer?
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The contents of a well-insulated tank are heated by a resistor of 2Ωin which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive.The rates of heat (Q), work (W) and change in internal energy (ΔU)during the process in kW area)b)c)d)Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The contents of a well-insulated tank are heated by a resistor of 2Ωin which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive.The rates of heat (Q), work (W) and change in internal energy (ΔU)during the process in kW area)b)c)d)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The contents of a well-insulated tank are heated by a resistor of 2Ωin which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive.The rates of heat (Q), work (W) and change in internal energy (ΔU)during the process in kW area)b)c)d)Correct answer is option 'A'. Can you explain this answer?.
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