Thermodynamics: free practice, theory and problems

1. Basic concepts

What is it about?

Thermodynamics is about energy: how it is stored, converted and flows as heat and work. A system is whatever you are studying (for example the gas inside a cylinder), and everything outside it is the surroundings. The system can be closed (only energy crosses the boundary), open (mass also flows in and out, as in a turbine) or isolated (neither energy nor mass crosses the boundary).

For an engineer this is the foundation for analyzing engines, refrigeration systems, power plants and heat pumps: they all rely on the fact that energy is never created or destroyed, only changes form.

Key quantities and formulas

• State properties describe the system exactly as it is now, independent of how it got there: pressure $p$, volume $V$, temperature $T$ and internal energy $U$.
• Ideal gas law: $pV = nRT$, with $R = 8.314$ J/(mol·K) and $T$ in kelvin.
• Temperature conversion: $T[\mathrm{K}] = t[^\circ\mathrm{C}] + 273.15$.
• Process types: isothermal ($T$ constant), isobaric ($p$ constant), isochoric ($V$ constant) and adiabatic ($Q = 0$, no heat exchange).
• First law for a closed system: $\Delta U = Q - W$, where $Q$ is the heat added and $W$ is the work done by the system on the surroundings.
• Work at constant pressure: $W = p\,\Delta V$.

How to solve the problems

1. Decide what kind of system it is (closed/open) and what kind of process (isothermal, isobaric, etc.).
2. Write down the known quantities in SI units, and remember to convert Celsius to kelvin.
3. Choose the right law: the ideal gas law for state properties, the first law for the energy balance.
4. Solve for the unknown, and check the sign: the work done by the system is positive in $\Delta U = Q - W$.

Example

2 mol of gas at a constant pressure of 150 kPa is heated from 300 K to 450 K. How much work does the gas do?

1. At constant pressure, $W = p\,\Delta V$, and from the ideal gas law $p\,\Delta V = nR\,\Delta T$.
2. $\Delta T = 450 - 300 = 150$ K.
3. $W = nR\,\Delta T = 2\cdot 8.314\cdot 150 \approx 2494$ J $\approx 2.49$ kJ.

Answer: The gas does about 2.49 kJ of work on the surroundings.

Common mistakes

• Using Celsius in the ideal gas law instead of kelvin.
• Mixing up the sign in the first law: $W$ is the work done by the system, not the work done on it.
• Believing that $\Delta U = 0$ holds in general, when it really only applies to isothermal processes for an ideal gas.
• Forgetting that pressure and volume must be in SI units (Pa and m³) when combined with $R = 8.314$ J/(mol·K).

Concepts in this part

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2. Heat and entropy

What is it about?

Heat is energy transferred because of a temperature difference. To understand how much heat is needed to warm something up, how efficiently a heat engine can convert heat into work, and why heat always flows from hot to cold, you need the concepts of heat capacity, entropy and Carnot efficiency.

This is central for engineers designing anything from cooling systems and heat pumps to steam turbines: the second law sets an upper limit on how much useful work you can extract from a given amount of heat.

Key quantities and formulas

• Sensible heat (temperature change without a phase change): $Q = mc\,\Delta T$, with specific heat capacity $c$ in kJ/(kg·K).
• Latent heat (phase change at constant temperature): $Q = mL$, where $L$ is the heat of fusion or vaporization.
• Enthalpy: $H = U + pV$. Useful for flow processes at constant pressure.
• Second law: the entropy of an isolated system can only increase or stay constant, $\Delta S_{isolated} \geq 0$.
• Entropy change at constant temperature: $\Delta S = Q/T$ (kelvin!).
• Carnot efficiency (the highest possible between two reservoirs): $\eta_{Carnot} = 1 - T_C/T_H$.
• Heat conduction through a wall (Fourier's law): $\dot Q = kA\,\Delta T/L$.

How to solve the problems

1. Decide whether the heat causes a temperature change (use $c$) or a phase change (use $L$), or both in sequence.
2. For multiple steps (e.g. heating then melting): compute the heat for each step and add them together.
3. For efficiency: identify $T_H$ and $T_C$ in kelvin, and use Carnot as an upper bound.
4. For entropy: make sure the temperature is in kelvin and that the sign matches the direction of the heat flow.

Example

How much heat is needed to warm 0.5 kg of ice from −5 °C to 0 °C and then melt it? ($c_{ice} = 2.1$ kJ/(kg·K), heat of fusion $L = 334$ kJ/kg)

1. Heating the ice: $Q_1 = mc\,\Delta T = 0.5\cdot 2.1\cdot 5 = 5.25$ kJ.
2. Melting: $Q_2 = mL = 0.5\cdot 334 = 167$ kJ.
3. Total: $Q = Q_1 + Q_2 = 5.25 + 167 = 172.25$ kJ.

Answer: About 172 kJ.

Common mistakes

• Using $Q = mc\,\Delta T$ across a phase change; the temperature is constant there, so you must use $Q = mL$ instead.
• Forgetting to convert to kelvin in the Carnot and entropy formulas.
• Believing that a heat engine can have an efficiency above the Carnot limit.
• Mixing up the heat of fusion and the heat of vaporization, which are very different values.

Concepts in this part

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3. Thermodynamic cycles

What is it about?

A thermodynamic cycle is a series of processes that returns the system to its starting point, so it can repeat cycle after cycle – like an engine or a refrigeration machine does. Idealized cycles (Otto, Diesel, Brayton, Rankine) are simplified models that let us calculate efficiency and work without knowing every detail of a real machine.

For an engineer this is the tool for comparing and improving power plants, engines, refrigeration systems and heat pumps: how much useful work or heat do you get out for each unit of energy you put in?

Key quantities and formulas

• Heat engine: takes in heat $Q_H$ from a hot reservoir, rejects $Q_C$ to a cold one, and delivers work $W = Q_H - Q_C$.
• Efficiency: $\eta = W/Q_H = 1 - Q_C/Q_H$.
• Heat pump/refrigerator: uses work $W$ to move heat from cold to hot. $COP_{heat} = Q_H/W$ and $COP_{cold} = Q_C/W$, with $COP_{heat} = COP_{cold} + 1$.
• Otto cycle (gasoline engine): heat is added at constant volume. Diesel cycle: heat is added at constant pressure.
• Brayton cycle (gas turbine/jet engine): compression, heat addition at constant pressure, expansion.
• Rankine cycle (steam power plant): evaporation, expansion in a turbine, condensation, pumping.
• Throttling (e.g. in an expansion valve): approximately constant enthalpy, but pressure and temperature drop.
• Isentropic process: reversible and adiabatic ($Q = 0$ and $\Delta S = 0$), used to idealize compressors and turbines.

How to solve the problems

1. Identify what kind of machine it is: a heat engine (delivers work) or a heat pump/refrigerator (uses work).
2. Set up the energy balance: $W = Q_H - Q_C$ for a heat engine.
3. Use $\eta$ or $COP$ to connect what you know with what you are looking for.
4. Check against the Carnot limit as a sanity check: real machines never exceed it.

Example

A heat pump has $COP_{heat} = 3.2$ and must deliver 8 kW of heat to a house. How much electric power does it need, and how much heat does it draw from the outside air?

1. $COP_{heat} = Q_H/W$, so $W = Q_H/COP_{heat} = 8/3.2 = 2.5$ kW.
2. Energy balance: $Q_C = Q_H - W = 8 - 2.5 = 5.5$ kW.

Answer: It uses 2.5 kW of electricity and draws 5.5 kW from the outside air.

Common mistakes

• Mixing up $COP_{heat}$ and $COP_{cold}$ – they are always different for the same machine.
• Believing a heat pump "creates" energy because $COP > 1$; it only moves heat and uses work to do so.
• Forgetting that $Q_H = Q_C + W$ must hold (energy conservation) in every cycle.
• Using Celsius instead of kelvin when computing the Carnot limit for $\eta$ or $COP$.

Concepts in this part

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4. Heat pumps and refrigeration

What is it about?

Heat flows by itself from hot to cold. A heat pump or a refrigerator moves heat the other way, from cold to hot, and pays for it with electrical work. The point is that you move much more heat than the electricity you use. That is why heat pumps are so popular in Norwegian homes.

Concepts and formulas

• Energy balance: the heat delivered on the hot side is the heat taken from the cold side plus the work.

• The coefficient of performance (COP) tells you how much useful heat you get per unit of electricity:

• The Carnot limit is the best possible COP between two temperatures. The temperatures must be in kelvin:

• The cycle: evaporator (absorbs heat outside or inside the fridge) → compressor (work in) → condenser (releases heat indoors) → expansion valve (pressure drops) → back to the evaporator.
• The larger the temperature difference, the lower the COP. That is why an air-to-air heat pump gives less per kWh of electricity on the coldest days.

How to solve the problems

1. Decide whether you are looking at heating ($Q_H$ is the benefit) or cooling ($Q_L$ is the benefit).
2. Use $\text{COP} = \text{benefit}/W$ and $Q_H = Q_L + W$.
3. For Carnot: convert to kelvin with $T = t + 273.15$ before calculating.

Example

A house needs 8 kW of heat. The heat pump has a COP of 3.2.

1. The electricity is $W = Q_H/\text{COP} = 8/3.2 = 2.5$ kW.
2. The heat taken from the outdoor air is $Q_L = Q_H - W = 8 - 2.5 = 5.5$ kW.
3. With an ordinary electric heater you would have used 8 kW of electricity. So you save 5.5 kW.

Common mistakes

• Using degrees Celsius in the Carnot formula. $20/(20 - 0)$ gives infinity, the correct value is $293.15/20 \approx 14.7$.
• Mixing up $\text{COP}_{HP}$ and $\text{COP}_{R}$. They differ by exactly 1.
• Thinking that a COP above 1 breaks energy conservation. The heat pump doesn't create energy, it moves it.

Concepts in this part

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