The First Law of Thermodynamics: Solved Problems

The first law of thermodynamics is simply conservation of energy written for engineers. It tells you that energy can move across a boundary as heat or work, and that whatever crosses must show up as a change in the energy stored inside. Once you choose your system carefully and stick to one sign convention, most first-law problems reduce to a short, tidy calculation, as the solved examples below show.

Stating the First Law

Closed systems

A closed system (a fixed mass, such as gas trapped in a piston-cylinder) exchanges heat and work but no mass. Using the common engineering convention where heat added to the system and work done by the system are positive:

$$ Q - W = \Delta U $$

Here \( Q \) is heat transfer, \( W \) is work and \( \Delta U \) is the change in internal energy, all in kJ. Kinetic and potential energy changes are usually negligible for a closed system at rest.

Open systems (steady flow)

Turbines, compressors, nozzles and heat exchangers have fluid flowing in and out. For steady flow with one inlet and one exit, the steady-flow energy equation (SFEE) is

$$ \dot{Q} - \dot{W} = \dot{m}\left[(h_2 - h_1) + \frac{v_2^2 - v_1^2}{2} + g(z_2 - z_1)\right] $$

where \( h \) is specific enthalpy (kJ/kg), \( \dot{m} \) is mass flow rate (kg/s) and \( \dot{Q}, \dot{W} \) are rates in kW. Enthalpy \( h = u + pv \) neatly packages the "flow work" needed to push fluid across the boundary.

Useful ideal-gas relations

For air treated as an ideal gas with constant specific heats:

• \( \Delta U = m c_v \Delta T \) and \( \Delta H = m c_p \Delta T \)
• \( c_p - c_v = R \), with \( c_p = 1.005 \), \( c_v = 0.718 \) and \( R = 0.287 \) kJ/(kg·K)
• Boundary work at constant pressure: \( W = p(V_2 - V_1) = mR(T_2 - T_1) \)

A Reliable Method for Every Problem

1. Sketch the system and draw its boundary.
2. Decide: closed or open? Steady or unsteady?
3. List the knowns with units and state assumptions (ideal gas, negligible KE, adiabatic, and so on).
4. Write the appropriate form of the first law and cancel zero terms.
5. Solve, then check signs, units and physical sense.

Solved Problem 1: Heating Air at Constant Pressure

A piston-cylinder contains 0.5 kg of air at 200 kPa and 300 K. Heat is added while the piston moves freely so the pressure stays constant, until the temperature reaches 450 K. Find the boundary work, the change in internal energy and the heat added.

Step 1: Boundary work. At constant pressure, \( pV = mRT \) gives \( p\Delta V = mR\Delta T \):

$$ W = mR(T_2 - T_1) = (0.5\ \text{kg})(0.287\ \text{kJ/kg·K})(450 - 300)\ \text{K} = 21.53\ \text{kJ} $$

Step 2: Change in internal energy.

$$ \Delta U = m c_v (T_2 - T_1) = (0.5\ \text{kg})(0.718\ \text{kJ/kg·K})(150\ \text{K}) = 53.85\ \text{kJ} $$

Step 3: Heat from the first law.

$$ Q = \Delta U + W = 53.85 + 21.53 = 75.38\ \text{kJ} $$

Sanity check. For a constant-pressure process, heat should equal the enthalpy change:

$$ m c_p \Delta T = (0.5)(1.005)(150) = 75.38\ \text{kJ} $$

The two answers agree. Also, the work is positive because the gas expands and pushes the piston out, and roughly 29% of the heat leaves again as work (\( R/c_p = 0.287/1.005 \approx 0.286 \)), which is the known fraction for a diatomic ideal gas.

Solved Problem 2: Power Output of a Steam Turbine

Steam enters a turbine at 2 kg/s with enthalpy 3230 kJ/kg and velocity 50 m/s. It leaves with enthalpy 2675 kJ/kg and velocity 120 m/s. Heat loss to the surroundings is 20 kW, and elevation change is negligible. Find the power output.

Step 1: Kinetic energy change per kilogram. Convert J/kg to kJ/kg by dividing by 1000:

$$ \frac{v_2^2 - v_1^2}{2} = \frac{(120)^2 - (50)^2}{2}\ \frac{\text{m}^2}{\text{s}^2} = \frac{14\,400 - 2500}{2} = 5950\ \text{J/kg} = 5.95\ \text{kJ/kg} $$

Step 2: Enthalpy change.

$$ h_2 - h_1 = 2675 - 3230 = -555\ \text{kJ/kg} $$

Step 3: Rearrange the SFEE. With \( \dot{Q} = -20 \) kW (heat leaves the turbine):

$$ \dot{W} = \dot{Q} - \dot{m}\left[(h_2 - h_1) + \Delta ke\right] $$

$$ \dot{W} = -20\ \text{kW} - (2\ \text{kg/s})\left[-555 + 5.95\right]\ \text{kJ/kg} = -20 + 1098.1 = 1078.1\ \text{kW} $$

The turbine delivers about 1.08 MW.

Sanity check. The kinetic energy term is only about 1% of the enthalpy drop, which is typical of turbines and explains why textbooks often neglect it. The positive answer confirms that work leaves the system, as expected for a turbine. A small industrial back-pressure turbine in a sugar mill or textile unit is of exactly this order of size.

Solved Problem 3: A Stirred Rigid Tank

A rigid, insulated tank holds 2 kg of air. A paddle wheel driven by an external motor does 30 kJ of work on the air. Find the temperature rise.

Step 1: Simplify. Rigid tank means no boundary work; insulated means \( Q = 0 \). Paddle work is done on the system, so \( W = -30 \) kJ.

Step 2: Apply the first law.

$$ 0 - (-30\ \text{kJ}) = m c_v \Delta T \quad \Rightarrow \quad \Delta T = \frac{30\ \text{kJ}}{(2\ \text{kg})(0.718\ \text{kJ/kg·K})} = 20.9\ \text{K} $$

Sanity check. Stirring work turns into internal energy and the temperature rises, which agrees with experience. The value is modest because 30 kJ is spread over 2 kg of gas.

Common Pitfalls

• Sign errors. Decide your convention at the start and write it down. Some books use \( \Delta U = Q + W \) with work done on the system positive; the physics is identical.
• Using \( c_p \) for \( \Delta U \). Internal energy always uses \( c_v \); enthalpy uses \( c_p \).
• Mixing J and kJ. Velocity terms come out in J/kg and must be divided by 1000 before adding to enthalpies.
• Treating a rigid tank as having boundary work. No volume change means no \( p\,dV \) work, although shaft or electrical work may still cross the boundary.

Did You Know?

• James Joule's paddle-wheel experiments in the 1840s showed that mechanical work and heat are interchangeable, which laid the groundwork for the first law.
• A "perpetual motion machine of the first kind", one that produces work from nothing, is ruled out by the first law; patent offices routinely reject such claims.
• Rudolf Clausius gave the first clear statement of the law in 1850.

Key Takeaways

• For closed systems use \( Q - W = \Delta U \); for steady flow use the SFEE with enthalpy.
• Draw the boundary first, then cancel the terms your assumptions make zero.
• Use \( c_v \) for internal energy and \( c_p \) for enthalpy, and keep J and kJ consistent.
• Cross-check answers with a second route, such as \( Q = m c_p \Delta T \) at constant pressure.

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