1.0 Concepts of Heat and Temperature
Calorimetry is the science of measuring the amount of heat energy absorbed or released during a physical or chemical process. To master this chapter, we must first clearly distinguish between Heat and Temperature, as they are often confused in daily language.
Key Differences
| Feature | Heat | Temperature |
|---|---|---|
| Definition | Form of energy that flows from hot to cold bodies. | Degree of hotness or coldness of a body. |
| S.I. Unit | Joule ($J$) | Kelvin ($K$) |
| Measurement | Measured by Calorimeter. | Measured by Thermometer. |
1.1 Thermal Capacity and Specific Heat Capacity
When you heat different substances, their temperatures rise at different rates. This is because every substance has a unique ability to store heat energy.
- Heat Capacity ($C'$): The amount of heat required to raise the temperature of the entire body by $1^\circ C$. Unit: $J/K$ or $J/^\circ C$.
- Specific Heat Capacity ($c$): The amount of heat required to raise the temperature of a unit mass ($1\,kg$) of a substance by $1^\circ C$. Unit: $J\,kg^{-1}\,K^{-1}$.
The Fundamental Heat Equation
$$Q = mc\Delta t$$
Where: $m$ = mass, $c$ = specific heat capacity, $\Delta t$ = change in temperature.
1.2 The Principle of Method of Mixtures
This is the "Golden Rule" of Calorimetry. When a hot body is mixed with a cold body, heat energy is transferred from the hot body to the cold body until they reach a final equilibrium temperature.
Heat lost by Hot Body = Heat gained by Cold Body
(Assuming no heat is lost to the surroundings)
Water has an exceptionally high specific heat capacity ($4200\,J\,kg^{-1}\,K^{-1}$). This means it can absorb a large amount of heat without a significant rise in its own temperature. This is why water is used in car radiators and for fomentation in hot water bags.
Calculate the heat energy required to raise the temperature of $2\,kg$ of water from $20^\circ C$ to $70^\circ C$. (Specific heat of water = $4200\,J\,kg^{-1}\,K^{-1}$)
Solution:
1. Identify Values: $m = 2\,kg$, $c = 4200\,J/kg\,K$, $\Delta t = (70 - 20) = 50^\circ C$.
2. Apply Formula: $Q = mc\Delta t$.
3. Calculation: $Q = 2 \times 4200 \times 50 = \mathbf{4,20,000\,J}$ (or $420\,kJ$).
Final Answer: $4.2 \times 10^5$ Joules of heat is required.
Land breeze and sea breeze occur because land has a lower specific heat capacity than water. Land heats up and cools down faster than the sea, creating the pressure differences that drive the coastal winds you feel at the beach!
2.0 Change of Phase and Latent Heat
When a substance changes from one state to another (e.g., ice to water), it absorbs or releases heat without any change in its temperature. This "hidden" heat is used to break the bonds between molecules rather than increasing their kinetic energy. In Physics, we call this Latent Heat.
Key Definitions
- Melting: Change from solid to liquid at a constant temperature (Melting Point).
- Vaporization: Change from liquid to gas at a constant temperature (Boiling Point).
- Specific Latent Heat ($L$): The amount of heat energy required to change the phase of unit mass of a substance without a change in temperature. Unit: $J/kg$.
2.1 Heating Curve of Water
A heating curve shows how temperature changes as heat is added. The flat plateaus on the graph represent the phase changes where temperature remains constant despite heat being supplied.
The Latent Heat Formula
$$Q = mL$$
Note: For ice, $L_{fusion} = 336,000\,J/kg$ (or $80\,cal/g$).
2.2 Natural Consequences of Latent Heat
These are high-frequency "Give Reason" questions in the ICSE Board Exam:
- Snow makes surroundings cold: When snow melts, it absorbs a huge amount of latent heat ($336\,J$ for every gram) from the surroundings, causing the temperature of the air to drop significantly.
- Steam burns are more severe than boiling water: Steam at $100^\circ C$ contains more heat energy than water at $100^\circ C$ because it possesses the Latent Heat of Vaporization ($2260\,J/g$).
- Bottled drinks are cooled more effectively by ice: Ice at $0^\circ C$ is better than water at $0^\circ C$ because ice absorbs additional heat from the drink equal to its latent heat of fusion to melt.
During phase change (Melting/Boiling), always remember that $\Delta t = 0$. You cannot use $Q = mc\Delta t$ for these parts of the process. You must use $Q = mL$.
How much heat is required to convert $10\,g$ of ice at $0^\circ C$ into water at $10^\circ C$?
Solution:
1. Step 1: Melting ($Q_1 = mL$)
$Q_1 = 0.01\,kg \times 336,000\,J/kg = 3360\,J$.
2. Step 2: Heating Water ($Q_2 = mc\Delta t$)
$Q_2 = 0.01\,kg \times 4200\,J/kg\,K \times 10 = 420\,J$.
3. Total Heat: $Q_1 + Q_2 = 3360 + 420 = \mathbf{3780\,J}$.
Final Answer: $3780$ Joules.
In very cold countries, farmers fill their fields with water to protect crops from frost. When water freezes into ice, it releases its latent heat of fusion to the surroundings, keeping the temperature of the crops from falling below $0^\circ C$.