Question:

Match the LIST-I with LIST-II
\begin{tabular}{|l|l|} \hline LIST-I & LIST-II
\hline A. If heat is added to the system & I. \(Q<0\)
B. If heat is removed from the system & II. \(W>0\)
C. If work is done on the system & III. \(Q>0\)
D. If work is done by the system & IV. \(W<0\)
\hline \end{tabular}
Choose the correct answer from the options given below:

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A simple way to remember the sign convention for \(\Delta U = Q - W\):

Energy IN (Heat added, Work done ON) increases internal energy.
Energy OUT (Heat removed, Work done BY) decreases internal energy.
Q is positive when it goes IN. W is positive when it comes OUT (as useful work).
Updated On: Sep 22, 2025
  • A - III, B - I, C - IV, D - II
  • A - I, B - III, C - II, D - IV
  • A - I, B - II, C - IV, D - III
  • A - III, B - IV, C - II, D - I
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
This question tests the standard sign conventions used in thermodynamics for heat (Q) and work (W) in the context of the first law, \(\Delta U = Q - W\).  
Step 2: Detailed Explanation: 
The most common convention in physics and engineering is as follows: 

A. If heat is added to the system: This increases the system's energy, so the heat transfer Q is considered positive. A matches III (\(Q>0\))
B. If heat is removed from the system: This decreases the system's energy, so the heat transfer Q is considered negative. B matches I (\(Q<0\))
C. If work is done on the system: The surroundings do work on the system (e.g., compression), which increases the system's energy. In the formula \(\Delta U = Q - W\), to make \(\Delta U\) increase, W must be negative. C matches IV (\(W<0\))
D. If work is done by the system: The system does work on the surroundings (e.g., expansion), which decreases the system's energy. In the formula \(\Delta U = Q - W\), to make \(\Delta U\) decrease, W must be positive. D matches II (\(W>0\))
Step 3: Final Answer: 
The correct matching is A-III, B-I, C-IV, D-II. 
 

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