Question
Download Solution PDFThe change in internal energy of a thermodynamical system which has absorbed 2 kcal of heat and done 400 J of work is (1 cal = 4.2 J)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
The change in internal energy (ΔU) of a thermodynamical system can be calculated using the first law of thermodynamics:
ΔU = Q - W
Where:
ΔU is the change in internal energy
Q is the heat absorbed by the system
W is the work done by the system
Given:
Heat absorbed, Q = 2 kcal = 2 × 4.2 × 103 J = 8400 J
Work done, W = 400 J
Now, applying the values in the equation:
ΔU = 8400 J - 400 J = 8000 J = 8 kJ
Thus, the change in internal energy is 8 kJ.
Last updated on Jun 6, 2025
-> AIIMS BSc Nursing Notification 2025 has been released.
-> The AIIMS BSc Nursing (Hons) Exam will be held on 1st June 2025 and the exam for BSc Nursing (Post Basic) will be held on 21st June 2025.
-> The AIIMS BSc Nursing Application Form 2025 can be submitted online till May 15, 2025.
-> AIIMS BSc Nursing Result 2025 Out Today at the official webiste of AIIIMS Portal.
-> The AIIMS BSc Nursing Cut Off 2025 has been Released by the AIIMS Delhi Along with the Result