Question:
Assertion (A): When \(1\,g\) of ice melts at constant temperature at a pressure of \(1\,atm\), the increase in internal energy is greater than \(80\,cal\).
Reason (R): During melting of ice, work is done on the ice.
Assertion (A): When \(1\,g\) of ice melts at constant temperature at a pressure of \(1\,atm\), the increase in internal energy is greater than \(80\,cal\).
Reason (R): During melting of ice, work is done on the ice.
Reason (R): During melting of ice, work is done on the ice.
Show Hint
Use sign convention carefully:
\[
Q=\Delta U+W
\]
where \(W\) is work done by the system.
Updated On: Jun 17, 2026
- Both A and R are true. R is the correct explanation of A
- Both A and R are true, but R is not the correct explanation of A
- A is true, R is false
- A is false, R is true
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
First law of thermodynamics:
\[
Q=\Delta U+W
\]
where:
\[
Q=\text{heat supplied}
\]
\[
\Delta U=\text{change in internal energy}
\]
\[
W=\text{work done by system}
\]
Step 1: Heat absorbed during melting. Latent heat of fusion: \[ Q=80\,cal \]
Step 2: Understand work interaction. During melting, volume decreases slightly for ice-water transition. Hence external work is done on the system. Therefore: \[ W<0 \] So, \[ \Delta U=Q-W \] Since \(W\) is negative, \[ \Delta U>Q \] Thus: \[ \Delta U>80\,cal \] Hence both Assertion and Reason are true and Reason correctly explains Assertion.
Step 1: Heat absorbed during melting. Latent heat of fusion: \[ Q=80\,cal \]
Step 2: Understand work interaction. During melting, volume decreases slightly for ice-water transition. Hence external work is done on the system. Therefore: \[ W<0 \] So, \[ \Delta U=Q-W \] Since \(W\) is negative, \[ \Delta U>Q \] Thus: \[ \Delta U>80\,cal \] Hence both Assertion and Reason are true and Reason correctly explains Assertion.
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