Introduction
Heat and Thermodynamics is a fundamental topic in JEE Advanced Physics. It deals with the concepts of heat, work, and energy, and their interconversion. Understanding this topic is crucial for solving problems related to the behavior of gases, changes in temperature, and various thermodynamic processes.
Temperature and Heat
Temperature
- Temperature is a measure of the average kinetic energy of the particles in a substance.
- It is measured in degrees Celsius (°C), Kelvin (K), or Fahrenheit (°F).
Heat
- Heat is a form of energy transfer between bodies due to a temperature difference.
- It is measured in joules (J) or calories (cal).
Relationship between Heat and Temperature
- When heat is added to a substance, its temperature increases if the substance is not undergoing a phase change.
- The amount of heat required to change the temperature of a substance is given by:
$$ Q = mc\Delta T $$
where:
- ( Q ) is the heat added,
- ( m ) is the mass,
- ( c ) is the specific heat capacity, and
- ( \Delta T ) is the change in temperature.
Example Calculation: Calculate the heat required to raise the temperature of 2 kg of water from 25°C to 75°C. (Specific heat capacity of water, ( c = 4200 , \text{J/kg·K} ))
$$ Q = mc\Delta T = 2 \times 4200 \times (75 - 25) = 420,000 , \text{J} $$
Thermodynamic Systems
Types of Systems
- Isolated System: No exchange of heat, work, or matter with surroundings.
- Closed System: Exchange of heat and work but not matter.
- Open System: Exchange of heat, work, and matter with surroundings.
State Variables
- Pressure (P)
- Volume (V)
- Temperature (T)
- Internal Energy (U)
Laws of Thermodynamics
Zeroth Law of Thermodynamics
- If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
- This law forms the basis for temperature measurement.
First Law of Thermodynamics
- Energy cannot be created or destroyed, only transferred or converted from one form to another.
- Mathematically, it is expressed as:
$$ \Delta U = Q - W $$
where:
- ( \Delta U ) is the change in internal energy,
- ( Q ) is the heat added to the system,
- ( W ) is the work done by the system.
Remember that work done by the system is taken as positive, and work done on the system is taken as negative.
Common MistakeA common mistake is to confuse the signs of ( Q ) and ( W ). Always follow the sign convention strictly.
Second Law of Thermodynamics
- Heat cannot spontaneously flow from a colder body to a hotter body.
- Entropy of an isolated system always increases over time.
- It introduces the concept of entropy (( S )):
$$ \Delta S \geq 0 $$
Third Law of Thermodynamics
- As temperature approaches absolute zero, the entropy of a perfect crystal approaches zero.
Thermodynamic Processes
Isothermal Process
- Temperature remains constant (( \Delta T = 0 )).
- For an ideal gas:
$$ PV = \text{constant} $$
ExampleExample: For an isothermal expansion of an ideal gas from volume ( V_1 ) to ( V_2 ):
$$ W = nRT \ln \left( \frac{V_2}{V_1} \right) $$
Adiabatic Process
- No heat exchange (( Q = 0 )).
- For an ideal gas:
$$ PV^\gamma = \text{constant} $$
where ( \gamma ) is the adiabatic index (( \gamma = \frac{C_p}{C_v} )).
Isobaric Process
- Pressure remains constant (( \Delta P = 0 )).
- Work done by the gas:
$$ W = P \Delta V $$
Isochoric Process
- Volume remains constant (( \Delta V = 0 )).
- No work is done (( W = 0 )).
Heat Engines and Refrigerators
Heat Engines
- Devices that convert heat energy into work.
- Efficiency (( \eta )) is given by:
$$ \eta = \frac{W_{\text{out}}}{Q_{\text{in}}} = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}} $$
Refrigerators
- Devices that transfer heat from a colder body to a hotter body.
- Coefficient of Performance (COP) is given by:
$$ \text{COP} = \frac{Q_{\text{cold}}}{W_{\text{input}}} $$
NoteThe efficiency of a heat engine is always less than 1, and the COP of a refrigerator is always greater than 1.
Carnot Cycle
- A theoretical cycle that is the most efficient.
- Consists of two isothermal and two adiabatic processes.
- Efficiency of a Carnot engine:
$$ \eta_{\text{Carnot}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} $$
ExampleExample: For a Carnot engine operating between 500 K and 300 K:
$$ \eta_{\text{Carnot}} = 1 - \frac{300}{500} = 0.4 \text{ or } 40% $$
Conclusion
Understanding Heat and Thermodynamics is crucial for solving problems in JEE Advanced Physics. Mastering the laws, processes, and cycles will enable you to tackle a wide range of questions effectively. Always remember to apply the correct formulas and sign conventions to avoid common mistakes.