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In a closed insulated container a liquid is stirred with a paddle to increase the temperature which of the following is true
Options
(a) ΔE = w ≠ 0, q = 0
(b) ΔE = w = q ≠ 0
(c) ΔE = 0, w = q ≠ 0
(d) w = 0, ΔE = q ≠ 0
Correct Answer:
ΔE = w ≠ 0, q = 0
Explanation:
The mathematical form of first law of thermodynamics : q = ΔE + W, Since the system is closed and insulated, q = 0, Paddle work is done on system. Therefore W ≠ 0. Temperature and hence internal energy of the system increases. Therefore ΔE ≠ 0.
Related Questions: - M⁺³ ion loses 3e⁻. Its oxidation number will be
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- M⁺³ ion loses 3e⁻. Its oxidation number will be
- Neutral divalent carbon species released as reaction intermediate in reaction
- Which of the following element,has the same oxidation state in all of its com.
- An example for a double salt is
- What is formed when hydrogen atom of the -OH group in phenol is replaced by
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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