The first law for closed systems

Energy cannot be created or destroyed; it can only change form or move. This statement, the first law of thermodynamics, is so familiar that it is easy to underestimate. It cannot be proved from anything simpler. It rests on a vast body of experiment, starting with Joule's measurements in the 1840s, in which falling weights stirred water in an insulated vessel and the water warmed by an amount that depended only on the work done, not on how it was done. No experiment since has contradicted it. In this lesson we write the first law for closed systems, where a fixed mass of matter exchanges energy with its surroundings only as heat and work, and apply it to the tanks, cylinders and vessels of the plant.

The energy balance

For any system undergoing any process, the net change in the system's total energy equals the net energy transferred across its boundary:

For a closed system, energy crosses the boundary only as heat and work. With the sign convention of Lesson 2 (heat in and work out positive), and writing \(Q\) and \(W\) for the net heat and work, the balance becomes

Most closed systems in engineering, such as a gas in a cylinder or water in a tank, do not move or rise appreciably, so \(\Delta KE = \Delta PE = 0\), and the balance reduces to \(Q - W = \Delta U = m(u_2 - u_1)\). Here \(W\) includes every kind of work: boundary work, shaft work from a stirrer, electrical work through a heater. It is often clearer to write the balance with explicit directions, such as \(Q_{\text{in}} + W_{e,\text{in}} - W_{b,\text{out}} = \Delta U\), using positive magnitudes throughout. Per unit mass the balance is \(q - w = \Delta u\), and as a rate it is \(\dot{Q} - \dot{W} = dE/dt\).

For a cycle, the initial and final states are the same, so \(\Delta E = 0\) and the net heat equals the net work: \(Q_{\text{net}} = W_{\text{net}}\). A heat engine operating in a cycle can deliver net work only by taking in more heat than it rejects. The first law says nothing about how much of the heat can become work; that limit comes from the second law, in Lesson 10.

A procedure for closed-system problems

Elena's procedure for Tomás has six steps. Sketch the system and mark the boundary. List the known properties at each state and the process (constant pressure, constant volume, adiabatic, and so on). Decide how each property will be found: tables for water and refrigerants, the ideal-gas relations for gases. Write the energy balance in its general form, then cancel the terms that are zero. Evaluate the boundary work along the actual process path. Finally, solve, and check that the answer has the right sign and a sensible size.

Rigid tanks: no boundary work

When the volume of a closed system is fixed, no boundary work is done. If there is no other work, the heat transfer equals the change in internal energy directly: \(Q = m(u_2 - u_1)\). For steam and water, internal energy may be read from tables or found from enthalpy as \(u = h - Pv\).

Constant-pressure processes and enthalpy

When a closed system expands or contracts at constant pressure, the boundary work is \(W_b = P(V_2 - V_1)\). Substituting this into the energy balance with no other work gives

So for a quasi-equilibrium constant-pressure process with only boundary work, the heat transfer equals the change in enthalpy. This is a second reason for defining enthalpy, and it is why the heat of a constant-pressure process can be read directly from the steam tables. If other work, such as an electric heater, is also present, it simply appears alongside: \(Q_{\text{in}} + W_{\text{other,in}} = H_2 - H_1\).

Ideal-gas processes

For an ideal gas, \(\Delta u = c_v\Delta T\) for any process, so closed-system problems reduce to finding the end temperatures and the boundary work. The program below takes 0.10 kg of air at 100 kPa and 300 K through five different processes, four of which double its volume, and applies the first law to each.

# First law for a closed system: Q - W = dU, for air (ideal gas, constant specific heats)
import math
R, cv = 0.287, 0.718                     # kJ/(kg.K)
m, P1, T1 = 0.10, 100.0, 300.0           # kg, kPa, K
V1 = m * R * T1 / P1                     # m3

def report(name, P2, V2, W):
    T2 = P2 * V2 / (m * R)
    dU = m * cv * (T2 - T1)
    Q = dU + W                           # first law: Q - W = dU
    print(f"{name:22s} T2 = {T2:6.1f} K  W = {W:7.3f} kJ  dU = {dU:7.3f} kJ  Q = {Q:7.3f} kJ")

print(f"initial state: P1 = {P1} kPa, T1 = {T1} K, V1 = {V1:.5f} m3")
V2 = 2 * V1                                            # expansions to double the volume
report("constant pressure", P1, V2, P1 * (V2 - V1))
report("isothermal", P1 * V1 / V2, V2, P1 * V1 * math.log(V2 / V1))
n = 1.3
P2 = P1 * (V1 / V2)**n
report("polytropic n = 1.3", P2, V2, (P2 * V2 - P1 * V1) / (1 - n))
k = 1.4
P2 = P1 * (V1 / V2)**k
report("adiabatic n = k = 1.4", P2, V2, (P2 * V2 - P1 * V1) / (1 - k))
report("constant volume, P x2", 2 * P1, V1, 0.0)

initial state: P1 = 100.0 kPa, T1 = 300.0 K, V1 = 0.08610 m3
constant pressure      T2 =  600.0 K  W =   8.610 kJ  dU =  21.540 kJ  Q =  30.150 kJ
isothermal             T2 =  300.0 K  W =   5.968 kJ  dU =   0.000 kJ  Q =   5.968 kJ
polytropic n = 1.3     T2 =  243.7 K  W =   5.388 kJ  dU =  -4.044 kJ  Q =   1.344 kJ
adiabatic n = k = 1.4  T2 =  227.4 K  W =   5.212 kJ  dU =  -5.216 kJ  Q =  -0.004 kJ
constant volume, P x2  T2 =  600.0 K  W =   0.000 kJ  dU =  21.540 kJ  Q =  21.540 kJ

Several lessons hide in this table. At constant pressure the air does 8.6 kJ of work but needs 30.2 kJ of heat, because its temperature doubles. In the isothermal expansion all the heat added leaves as work, since \(\Delta U = 0\). In the adiabatic expansion the work comes entirely from the internal energy, so the temperature falls to 227 K; the tiny nonzero \(Q\) is a rounding effect, because the table value \(c_v = 0.718\) is not exactly \(R/(k - 1) = 0.7175\). The polytropic expansion with \(n = 1.3\) lies between: some heat enters, but not enough to keep the temperature up. And heating at constant volume raises the temperature as much as constant-pressure heating does, with 29 percent less heat, because no work is done.

These results show why \(c_p\) exceeds \(c_v\). Raising 0.10 kg of air by 300 K took 30.15 kJ at constant pressure and 21.54 kJ at constant volume. Dividing by \(m\Delta T = 30\) gives 1.005 and 0.718 kJ/(kg·K): exactly the tabulated \(c_p\) and \(c_v\).

Work that is not boundary work

Paddle wheels, electric heaters and other devices can transfer energy as work even when the boundary does not move. Such work is always done on the system, never by it; you cannot run a paddle wheel backwards by cooling the water. That one-way character is an early hint of the second law.

Specific heats of liquids in closed systems

For incompressible substances, \(\Delta U = mc\Delta T\), and because the volume barely changes, boundary work is negligible. Heating or cooling a liquid or a solid therefore needs heat (or non-boundary work) equal to \(mc\Delta T\). The same relation governs how long a cold boiler takes to warm up: its 40 tonnes of steel, with \(c = 0.46\ \text{kJ/(kg·K)}\), need 18,400 kJ per degree, in addition to the water inside. Warming steel too fast causes thermal stresses, so start-up rates are limited by the boiler manufacturer, typically to a few degrees per minute.

Common pitfalls

Most errors in closed-system problems come from three sources. The first is using the wrong property: enthalpy where internal energy is needed, or vice versa. In a closed system, the energy stored is \(U\); enthalpy appears only because constant-pressure boundary work combines neatly with it. The second is forgetting boundary work when the volume changes, or including it when the volume is fixed. The third is sign confusion. Writing the balance with explicit in and out terms and positive magnitudes, then checking whether the system's energy should rise or fall, catches almost all of them.