Question

The given graph represents the cooling curve of a liquid.

(a) State the freezing temperature of the liquid.
(b) Name the phase change happening at the region QR.
(c) In which state (solid / liquid) does the above substance liberate heat at a faster rate? Justify.

Hint:

On a cooling curve, the slope tells you the rate of cooling. Steeper slope = faster cooling. A flat plateau = phase change at a constant temperature.

Answer 1

Step 1: Understanding the Concept:
During a phase change (like freezing), the temperature of a substance remains constant even though it is losing heat.
Step 2: Detailed Explanation:
On a cooling curve (temperature vs. time graph), a phase change is represented by a horizontal plateau (a line with zero slope).
In the provided graph, the segment QR is horizontal.
The temperature corresponding to this horizontal part is \( 20^\circ\text{C} \).
This indicates that the liquid is freezing into a solid at this constant temperature.
Step 3: Final Answer:
The freezing temperature is \( 20^\circ\text{C} \).
(b)
Step 1: Understanding the Concept:
Cooling involves the removal of heat, leading to a transition from a higher energy state (liquid) to a lower energy state (solid).
Step 2: Detailed Explanation:
The curve starts from a higher temperature (liquid phase PQ).
The region QR is the transition period where temperature is constant while heat is being released (Latent heat of fusion).
Since the substance is being cooled and the temperature is constant, it is changing from the liquid state to the solid state.
This process is known as freezing or solidification.
Step 3: Final Answer:
The phase change is freezing.
(c)
Step 1: Understanding the Concept:
The rate of cooling (rate of temperature drop) is indicated by the slope of the cooling curve.
Step 2: Detailed Explanation:
The slope of the curve represents how fast the temperature is changing over time (\( \Delta T / \Delta t \)).
- Region PQ (Liquid): Temperature drops from \( 45^\circ\text{C} \) to \( 20^\circ\text{C} \) in 20 seconds. Slope = \( 25/20 = 1.25^\circ\text{C/s} \).
- Region RS (Solid): Temperature drops from \( 20^\circ\text{C} \) to \( 0^\circ\text{C} \) in 5 seconds (from 30s to 35s). Slope = \( 20/5 = 4^\circ\text{C/s} \).
Since the slope is much steeper in the RS region, the temperature falls much faster in the solid state.
Assuming heat is lost to the same environment, a faster temperature drop implies heat is being liberated at a faster rate or the specific heat is lower.
Step 3: Final Answer:
The substance liberates heat at a faster rate in the solid state because the slope of the curve is steeper in region RS compared to PQ.