| ⇦ | 
| ⇨ |
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: - Heat of combustion ΔH for C(s) , H₂(g)and CH₄(g) are -94, -68 and -213 kcal/mol.
- The elements with atomic number 9,17,35,53,85 are called
- dsp² hybridisation is found in
- The amount of electricity requried to deposit 0.9 g of aluminium, when the electrode
- Which one of the following contain isopropyl group
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Heat of combustion ΔH for C(s) , H₂(g)and CH₄(g) are -94, -68 and -213 kcal/mol.
- The elements with atomic number 9,17,35,53,85 are called
- dsp² hybridisation is found in
- The amount of electricity requried to deposit 0.9 g of aluminium, when the electrode
- Which one of the following contain isopropyl group
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply