Heating and Cooling Curves

Heating and cooling curves show how the temperature of a substance changes over time as heat is added or removed. These graphs help visualize phase changes (melting, boiling, freezing, condensing) and the energy involved in these processes.

Key Regions of a Heating Curve

Region A–B (Solid heating): Temperature increases as solid absorbs heat, increasing kinetic energy.
Region B–C (Melting): Phase change from solid to liquid; temperature remains constant while potential energy increases.
Region C–D (Liquid heating): Temperature increases as liquid absorbs heat, increasing kinetic energy.
Region D–E (Boiling): Phase change from liquid to gas; temperature remains constant while potential energy increases.
Region E–F (Gas heating): Temperature increases as gas absorbs heat.

Cooling curves are simply the reverse of heating curves. They show energy being released instead of absorbed, with phase changes occurring at the same temperatures but in reverse order.

Key Equations Used

During temperature change (no phase change):
\[ Q = mc\Delta T \]
During phase change (no temperature change):
\[ Q = mL \] where \( L \) is the specific latent heat (fusion or vaporization)

Important Notes

Temperature does not change during a phase change.
More heat is needed for vaporization than melting (latent heat of vaporization > latent heat of fusion).
Slope of each section depends on the specific heat capacity of that phase.

Calculate the total energy needed to convert 100 g of ice at 0°C to steam at 100°C.
Given:

Specific heat of water: \( c = 4.18 \, \text{J/g°C} \)
Latent heat of fusion: \( L_f = 334 \, \text{J/g} \)
Latent heat of vaporization: \( L_v = 2260 \, \text{J/g} \)

Solution:

Step 1: Melt the ice (0°C) → \( Q_1 = mL_f = 100 \times 334 = 33,400 \, \text{J} \)
Step 2: Heat the water from 0°C to 100°C → \( Q_2 = mc\Delta T = 100 \times 4.18 \times 100 = 41,800 \, \text{J} \)
Step 3: Vaporize the water → \( Q_3 = mL_v = 100 \times 2260 = 226,000 \, \text{J} \)

Total heat energy:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 33,400 + 41,800 + 226,000 = 301,200 \, \text{J} \]

How much energy is released when 50 g of steam at 100°C is cooled and frozen into ice at 0°C?
Use the same values as Example 1.

Solution:

Step 1: Condense the steam → \( Q_1 = mL_v = 50 \times 2260 = 113,000 \, \text{J} \)
Step 2: Cool water from 100°C to 0°C → \( Q_2 = mc\Delta T = 50 \times 4.18 \times 100 = 20,900 \, \text{J} \)
Step 3: Freeze the water → \( Q_3 = mL_f = 50 \times 334 = 16,700 \, \text{J} \)

Total energy released:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 113,000 + 20,900 + 16,700 = 150,600 \, \text{J} \]

Written by Thenura Dilruk