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

(a) Freezing Temperature:
The freezing temperature is the temperature at which the liquid turns into a solid. During this phase change, the temperature remains constant even as heat is being liberated. The horizontal section of the graph, QR, represents this phase change. The temperature corresponding to this section is 20 \(^\circ\)C.
(b) Phase Change at QR:
The substance is cooling. At the constant temperature region QR, the liquid is changing its state to solid. This process is called freezing or solidification.
(c) State with faster heat liberation:
The substance liberates heat at a faster rate in the liquid state (region PQ).
Justification: The rate of heat liberation is proportional to the rate of cooling. The rate of cooling is represented by the slope (steepness) of the temperature-time graph. The slope of the graph in the region PQ (liquid state) is steeper than the slope in the region RS (solid state). A steeper slope means a larger change in temperature per unit time, indicating a faster rate of heat loss.