Heating Curve
Table of Contents
A heating curve is a graph that shows how the temperature of a substance changes as heat is added. It helps us understand what happens when a substance is heated, including both temperature changes and phase changes such as melting and boiling. [1,3,5]
Heating curves are commonly used to study pure substances like water. Understanding these curves benefits both theoretical learning and practical experiments.
Graphical Representation
As heat is added, a substance may first get warmer and then change state when it reaches its melting or boiling point. The heating curve shows where the added heat is used to alter the physical state. For a pure substance at constant pressure, these changes occur at fixed temperatures. These fixed-temperature regions appear as horizontal plateaus on the heating curve. [1,3–5]
The key features of a heating curve are the following:
- The vertical axis shows temperature.
- The horizontal axis shows the heat added or time taken.
- Sloped lines show the temperature change.
- Flat lines show the phase change.
Often, the horizontal axis may also represent time, provided heat is supplied at a constant rate.
Examples
Heating Curve of Water [1,3–5]
The heating curve of water is one of the most common examples. It shows how ice changes into liquid water and then into steam as heat is added.
The curve has five main parts.
| Part of Curve | State or Change | Temperature Behavior |
|---|---|---|
| Ice warming | Solid ice | Temperature increases |
| Melting | Solid ice changes to liquid water | Temperature remains at 0°C |
| Liquid water warming | Liquid water | Temperature increases |
| Boiling | Liquid water changes to steam | Temperature remains at 100°C |
| Steam warming | Water vapor | Temperature increases |
The melting point of water is 0°C, and its boiling point is 100°C at 1 atm pressure. These values apply to pure water under normal atmospheric pressure. If the pressure changes, the boiling point can also change. For mixtures or impure substances, phase changes may occur over a temperature range instead of at a sharp plateau.
During the sloped parts of the curve, the added heat increases the kinetic energy of particles, so the temperature rises. During flat regions, the added heat is used to overcome intermolecular forces and change the substance’s state. Therefore, the temperature remains constant during melting and boiling, even though heat is still being absorbed. [2,5]
Equations
Different parts of a heating curve require different heat equations. [2]
When Temperature Changes
For sloped regions, temperature changes. The heat absorbed is calculated using:
q = mcΔT
Where:
q: Heat absorbed
m: Mass of the substance
c: Specific heat capacity
ΔT: Change in temperature
This equation is used when a substance is heated while remaining in a single physical state, such as warming ice, liquid water, or steam.
When Phase Changes
For flat regions, the temperature remains constant while the state changes. The heat absorbed is calculated using:
q = mL
Where:
q: Heat absorbed
m: Mass of the substance
L: Latent heat
For melting, use the latent heat of fusion, Lf.
For boiling, use the latent heat of vaporization, Lv.
Calculating Total Heat from a Heating Curve
To calculate the total heat absorbed during a heating process, divide the curve into separate parts.
Follow these steps:
- Identify each segment of the heating curve.
- Decide whether each segment is sloped or flat.
- Use q = mcΔT for sloped regions.
- Use q = mL for phase-change regions.
- Add the heat values from all segments.
For example, heating ice below 0°C until it becomes liquid water above 0°C involves three steps:
- Heating ice to 0°C
- Melting ice at 0°C
- Heating liquid water above 0°C
Each step must be calculated separately because the energy is used differently in each region.
Heating Curve vs. Cooling Curve
A heating curve shows what happens when heat is added to a substance. A cooling curve is the reverse of a heating curve. It shows what happens when heat is removed. [1,6]
| Feature | Heating Curve | Cooling Curve |
|---|---|---|
| Direction of change | The substance may change from solid → liquid → gas | The substance may change from gas → liquid → solid |
| Energy transfer | Heat is absorbed by the substance | Heat is released by the substance |
| Phase-change temperatures | For a pure substance at fixed pressure, melting occurs at the melting point and boiling occurs at the boiling point | For a pure substance at fixed pressure, condensation occurs at the boiling point, and freezing occurs at the melting point |
| Temperature behavior during phase change | Temperature remains constant while heat is used to change state | Temperature remains constant while heat is released during change of state |
| Example | Ice melts into water and then boils into steam | Steam condenses into water and then freezes into ice |
Applications of Heating Curves
- Identifying melting/boiling points from a graph [2,6]
- Calculating heat absorbed during calorimetry
- Comparing substances by their specific heat or latent heat
- Explaining why the temperature remains constant during melting and boiling
- Comparing the heat required for different segments of a heating process
Solved Problems
Problem 1: How much heat is needed to raise the temperature of 50 g of water from 25°C to 75°C? The specific heat capacity of water is 4.18 J/g°C.
Answer
Given:
m = 50 g
c = 4.18 J/g°C
ΔT = 75°C – 25°C = 50°C
From the heat equation:
q = mcΔT
=> q = 50 × 4.18 × 50
=> q = 10,450 J
Therefore, 10,450 J of heat is required.
Problem 2: How much heat is needed to melt 20 g of ice at 0°C? The latent heat of fusion of ice is 334 J/g.
Answer
Given:
m = 20 g
Lf = 334 J/g
From the heat equation:
q = mLf
=> q = 20 × 334
=> q = 6,680 J
Therefore, 6,680 J of heat is needed to melt the ice.
Problem 3: How much heat is required to convert 10 g of ice at -10°C into water at 25°C?
Use:
Specific heat capacity of ice = 2.1 J/g°C
Latent heat of fusion of ice = 334 J/g
Specific heat capacity of water = 4.18 J/g°C
Answer
Step 1: Heat ice from -10°C to 0°C
q = mcΔT
=> q = 10 × 2.1 × 10
=> q = 210 J
Step 2: Melt ice at 0°C
q = mLf
=> q = 10 × 334
=> q = 3,340 J
Step 3: Heat water from 0°C to 25°C
q = mcΔT
q = 10 × 4.18 × 25
q = 1,045 J
Total heat:
qtotal = 210 + 3,340 + 1,045
=> qtotal = 4,595 J
Therefore, 4,595 J of heat is required.
