Question
Download Solution PDFA closed system undergoes a process in which the work done by the system is 100 J and the internal energy decreases by 50 J. According to the first law of thermodynamics, what is the amount of heat transferred into or out of the system?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The first law of thermodynamics
This law states that the total heat given or taken out of the system will always be equal to the work done and change in the internal energy of the system.
⇒ Q = W + ΔU
Where Q = heat, W = work done, and ΔU = change in internal energy
Calculation:
Given W = 100 J, and ΔU = -50 J
By the first law of thermodynamics,
⇒ Q = W + ΔU
⇒ Q = 100 + (-50)
⇒ Q = 50 J
Last updated on Jul 1, 2025
-> SSC JE ME Notification 2025 has been released on June 30.
-> The SSC JE Mechanical engineering application form are activated from June 30 to July 21.
-> SSC JE 2025 CBT 1 exam for Mechanical Engineering will be conducted from October 2 to 31.
-> SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.