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

Step 1: Understanding the Concept:
Phase change occurs at a constant temperature (plateaus in the T-Q graph).
Temperature change results in a sloped line. The slope \(dT/dQ = \frac{1}{mc}\) depends on specific heat capacity.
Step 2: Detailed Explanation:
1. Ice heating (-20 to 0°C): Slope is positive.
2. Melting (0°C): Horizontal line (Latent heat of fusion).
3. Water heating (0 to 100°C): Positive slope (half the slope of ice as \(s_w>s_{ice}\)).
4. Boiling (100°C): Horizontal line (Latent heat of vaporization). This line is much longer than melting because \(L_v \gg L_f\).
5. Steam heating (above 100°C): Positive slope.
Graph (1) correctly shows these stages with appropriate plateaus at \(0^\circ C\) and \(100^\circ C\).
Step 4: Final Answer:
Graph (1) is the correct representation.