Concept:
When heat is supplied to water over a wide temperature range, the temperature does not always increase uniformly.
At certain temperatures, heat is absorbed without any rise in temperature due to {change of phase}.
These appear as horizontal (flat) portions in a temperature vs heat graph.
Key phase changes for water:
Ice melts at \(0^\circ\text{C}\)
Water boils at \(100^\circ\text{C}\)
Step 1: Heating ice from \(-20^\circ\text{C}\) to \(0^\circ\text{C}\)
In this region:
Water is in solid (ice) form
Temperature increases linearly with heat supplied
Hence, the graph must show a {sloping straight line} from \(-20^\circ\text{C}\) to \(0^\circ\text{C}\).
Step 2: Melting of ice at \(0^\circ\text{C}\)
At \(0^\circ\text{C}\):
Ice changes to water
Temperature remains constant
Heat supplied is used as latent heat of fusion
Thus, the graph must have a {horizontal segment at \(0^\circ\text{C}\)}.
Step 3: Heating water from \(0^\circ\text{C}\) to \(100^\circ\text{C}\)
Now:
Water is in liquid state
Temperature rises uniformly with heat
So the graph again shows a {sloping straight line} up to \(100^\circ\text{C}\).
Step 4: Boiling of water at \(100^\circ\text{C}\)
At \(100^\circ\text{C}\):
Water changes to steam
Temperature remains constant
Heat supplied is used as latent heat of vaporization
Hence, there must be another {horizontal segment at \(100^\circ\text{C}\)}.
Step 5: Heating steam from \(100^\circ\text{C}\) to \(120^\circ\text{C}\)
Finally:
Steam temperature increases
Temperature again rises linearly with heat
This gives the final sloping segment beyond \(100^\circ\text{C}\).
Step 6: Identify the correct graph
The correct graph must therefore show:
Rising line from \(-20^\circ\text{C}\) to \(0^\circ\text{C}\)
Flat portion at \(0^\circ\text{C}\)
Rising line from \(0^\circ\text{C}\) to \(100^\circ\text{C}\)
Flat portion at \(100^\circ\text{C}\)
Rising line up to \(120^\circ\text{C}\)
Among the given options, Graph 4 satisfies all these conditions.