Question

Ice is heated from \(-20^\circ C\) to \(200^\circ C\). Which of the following temperature (T) vs heat (Q) graph is correct ?

• A.
• B.
• C.
• D.

Hint:

Vaporization always takes more energy than melting. Thus, the horizontal line at \(100^\circ C\) must always be significantly longer than the one at \(0^\circ C\).

Answer 1

The Correct Option is A

To determine the correct Temperature (T) vs Heat (Q) graph for heating ice from \(-20^\circ C\) to \(200^\circ C\), we need to consider the phase changes and temperature increases at each stage:

1. Initial Heating of Ice: The ice is initially at \(-20^\circ C\). As heat is added, the temperature of the ice increases linearly until it reaches \(0^\circ C\). This is depicted as an upward sloping line in the graph.
2. Melting of Ice: At \(0^\circ C\), the ice begins to melt. During this phase change, the temperature remains constant until all the ice has melted into water. This translates to a horizontal line on the graph.
3. Heating of Water: Once the ice has completely melted into water, the temperature of the water rises linearly with added heat until it reaches \(100^\circ C\).
4. Boiling of Water: At \(100^\circ C\), the water begins to boil. During the boiling phase, the temperature again remains constant as the water changes into steam. This is represented by another horizontal line.
5. Heating of Steam: Once all the water has turned into steam, further addition of heat will increase the temperature of the steam linearly from \(100^\circ C\) to \(200^\circ C\).

Now, considering these phases, the correct graph should display two horizontal sections where the phase changes occur (melting and boiling) and three linear sections of increasing slope where the temperature of ice, water, and steam rises.

The provided graph option that matches this description appears as shown below:

Thus, option (a) is the correct representation of the Temperature (T) vs Heat (Q) graph in this scenario.