First Law of Thermodynamics: Closed Systems, Specific Heats and Steady-Flow Devices

This chapter answers Section XE4.3 “First Law of Thermodynamics” of XE4 Thermodynamics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam: “Concept of energy and various forms of energy; Internal energy, enthalpy; Specific heats; First law applied to elementary processes, closed systems and control volumes, steady flow energy equation applied to simple engineering devices.” It teaches the energy balance with one stated sign convention, internal energy, enthalpy and the specific heats of an ideal gas (Mayer’s relation), the elementary processes, cycles, and the steady-flow energy equation for nozzles, diffusers, turbines, compressors, throttles, heat exchangers and mixing chambers. The numericals are isobaric heating, polytropic heat transfer, nozzle exit velocity, turbine power, compressor work and the final temperature of a tank being filled. Every property value is given in the stem.

1. Forms of energy and the first law for a closed system

Energy is stored in a system as kinetic energy KE = ½mV², potential energy PE = mgz and internal energy U, the sum of all microscopic energies (molecular translation, rotation, vibration, intermolecular and chemical bonding energy). The total is E = U + KE + PE. Energy crosses a boundary only as heat Q or work W (and, for an open system, with the mass that crosses it). U is a property and extensive, and so is E; the specific values u = U/m and e = E/m are intensive.

First law for a closed system. Convention used throughout: Q is positive when heat flows INTO the system, W is positive when the system does work ON its surroundings. Then for a process Q − W = ΔU + ΔKE + ΔPE, and for a cycle ∮δQ = ∮δW, since E returns to its initial value. Q and W separately depend on the path, but Q − W = ΔE does not: this is what makes E a property. In differential form δQ − δW = dE. Some textbooks take W as work done on the system and write Q + W = ΔE; the physics is identical, and the stem must state the convention.

⚠️ Sign check
A gas compressed with 200 kJ of work while 50 kJ of heat leaves it has Q = −50 kJ and W = −200 kJ, so ΔU = −50 − (−200) = +150 kJ. In a stationary system KE and PE changes are normally zero and ΔE = ΔU; do not drop them when the stem gives a moving or lifted system.

2. Enthalpy, specific heats and Mayer’s relation

Enthalpy is defined as H = U + pV, or h = u + pv per unit mass; it is a property. For a constant-pressure quasi-static process in a closed system with no other work, W = p ΔV and Q = ΔU + pΔV = ΔH. The specific heats are c_v = (∂u/∂T)_v and c_p = (∂h/∂T)_p. For an ideal gas, u and h depend on temperature only (Joule’s free-expansion experiment shows no temperature change for a low-density gas), so du = c_v dT and dh = c_p dT for any process of that gas, whatever the path.

For an ideal gas h = u + RT, so dh/dT − du/dT gives Mayer’s relation c_p − c_v = R (per unit mass; in molar terms C_p − C_v = R_u). With γ = c_p/c_v, c_v = R/(γ − 1) and c_p = γR/(γ − 1). Since c_p exceeds c_v (a constant-pressure heating also does expansion work), γ > 1. For air treat R = 287 J/kg·K and γ = 1.4, so c_v = 717.5 and c_p = 1004.5 J/kg·K. Monatomic gases have γ ≈ 1.67 and diatomic gases near room temperature γ ≈ 1.4; γ falls as temperature rises because internal modes are excited. For liquids and solids c_p ≈ c_v ≈ c and the single c is used.

⚠️ c_v is not only for constant volume
For an ideal gas ΔU = m c_v ΔT and ΔH = m c_p ΔT hold for every process, including isobaric and adiabatic ones. Writing Q = m c_v ΔT is valid only for a constant-volume process.

3. Elementary processes of an ideal gas

For a closed system of m kg of ideal gas, quasi-static, with W the work done BY the gas and Q the heat added, apply Q − W = ΔU with ΔU = m c_v (T₂ − T₁).

ProcessWork W by the gasHeat Q added
Isochoric0m c_v (T₂ − T₁)
Isobaricp (V₂ − V₁) = mR (T₂ − T₁)m c_p (T₂ − T₁)
IsothermalmRT ln(V₂/V₁) = mRT ln(p₁/p₂)Q = W (ΔU = 0)
Reversible adiabatic, pVᵞ = CmR (T₁ − T₂)/(γ − 1) = −ΔU0
Polytropic, pVⁿ = CmR (T₁ − T₂)/(n − 1)ΔU + W = W (γ − n)/(γ − 1)

For the polytropic result, Q = m c_v (T₂ − T₁) + mR (T₁ − T₂)/(n − 1) = mR (T₁ − T₂)[1/(n − 1) − 1/(γ − 1)] = W (γ − n)/(γ − 1). So when 1 < n < γ the gas expanding (W > 0) receives heat even though its temperature falls, because it does more work than the fall in its internal energy. Worked example. 2 kg of air (R = 0.287 kJ/kg·K, γ = 1.4) expands polytropically with n = 1.25 from 600 K to 500 K. W = 2 × 0.287 × 100/0.25 = 229.6 kJ; ΔU = 2 × 0.7175 × (−100) = −143.5 kJ; Q = 229.6 − 143.5 = +86.1 kJ, which agrees with 229.6 × 0.15/0.4 = 86.1 kJ.

⚠️ Adiabatic is not isothermal
In a reversible adiabatic process Q = 0 but T changes, because the work comes out of internal energy. The relation pVᵞ = C holds only for a reversible (quasi-static, frictionless) adiabatic process of an ideal gas with constant γ.

4. Closed-system processes and cycles

In a closed system the first law is a balance in three unknowns Q, W and ΔU, of which two are usually given. Heating at constant pressure of 0.5 kg of an ideal gas with R = 0.287 kJ/kg·K and γ = 1.4 from 300 K to 450 K: c_v = 0.7175, c_p = 1.0045 kJ/kg·K; ΔU = 0.5 × 0.7175 × 150 = 53.8 kJ; W = mRΔT = 0.5 × 0.287 × 150 = 21.5 kJ; Q = ΔU + W = 75.3 kJ, which equals m c_p ΔT = 0.5 × 1.0045 × 150 = 75.3 kJ. About 29% of the heat goes to work, which is 1 − 1/γ = R/c_p.

For a cycle ΔU = 0 over the whole cycle, so the net heat equals the net work: ΣQ = ΣW. A cycle that receives 900 kJ and rejects 600 kJ of heat produces 300 kJ of net work and, if it is an engine, has thermal efficiency 300/900 = 33.3%. For a series of processes, ΔU of each leg adds to zero over the cycle, a useful check that a missing term has been found.

🎯 Free expansion
A gas expanding into an evacuated, rigid, insulated space has Q = 0 and W = 0, so ΔU = 0. For an ideal gas this means the temperature is unchanged; for a real gas it generally changes slightly, which is why ideal gas is the idealisation, not a law.

5. Control volume and the steady-flow energy equation

For a control volume, mass is conserved: dm_cv/dt = Σṁ_in − Σṁ_out. A unit mass entering or leaving has also to be pushed against the pressure, which requires flow work pv, so the energy carried per unit mass across the boundary is u + pv + V²/2 + gz = h + V²/2 + gz. This is why enthalpy, not internal energy, appears. For steady flow (no change with time inside the control volume, so ṁ_in = ṁ_out and E_cv constant), with Q̇ the rate of heat added and Ẇ the shaft power done by the fluid:

Q̇ − Ẇ = ṁ [ (h₂ − h₁) + (V₂² − V₁²)/2 + g (z₂ − z₁) ], or per unit mass q − w = Δh + ΔV²/2 + gΔz. With h in kJ/kg and V in m/s, convert V²/2 to kJ/kg by dividing by 1000. The equation is the first law, with the same sign convention; it holds with or without friction, for any fluid.

DeviceUsual simplificationResult
Nozzlew = 0, q ≈ 0, Δz ≈ 0V₂² = V₁² + 2 (h₁ − h₂)
Diffusersame as nozzle, V fallsh₂ = h₁ + (V₁² − V₂²)/2 > h₁
Turbineq small, Δz = 0w = (h₁ − h₂) + (V₁² − V₂²)/2 − q
Compressor, pumpq small or loss−w = (h₂ − h₁) + ΔV²/2 − q
Throttleq = w = 0, ΔV²/2 smallh₂ = h₁ (T unchanged for ideal gas)
Heat exchanger (insulated outside)w = 0, Σ over both streamsΣṁh in = Σṁh out
Mixing chamberw = q = 0ṁ₁h₁ + ṁ₂h₂ = (ṁ₁ + ṁ₂) h₃

6. Worked examples and a tank being filled

Nozzle. Air (c_p = 1.005 kJ/kg·K) enters an adiabatic nozzle at 600 K and 50 m/s and leaves at 500 K: V₂² = V₁² + 2 c_p (T₁ − T₂) × 1000 = 2500 + 2 × 1005 × 100 = 203 500 m²/s², so V₂ = 451 m/s. Turbine. Steam at 5 kg/s enters with h₁ = 3230 kJ/kg and V₁ = 50 m/s, leaves with h₂ = 2550 kJ/kg and V₂ = 150 m/s, and loses 40 kW as heat: Ẇ = ṁ[(h₁ − h₂) + (V₁² − V₂²)/2000] + Q̇ = 5 × [680 − 10] − 40 = 3310 kW, with Q̇ = −40 kW. The kinetic term is small beside Δh, a usual check.

Compressor. Air at 2 kg/s is compressed from 300 K to 440 K, with c_p = 1.005 kJ/kg·K, negligible kinetic energy and 30 kW lost as heat to the surroundings: −Ẇ = ṁ c_p ΔT − Q̇ = 2 × 1.005 × 140 + 30 = 311.4 kW of work input, with Q̇ = −30 kW. The heat loss adds to the work required.

Unsteady filling. A rigid, adiabatic, initially evacuated tank is connected to a supply line of ideal gas at a constant temperature T_in. For the tank as the control volume, Q = W = 0, kinetic and potential energies neglected, no outflow, and the initial energy is zero, so the energy balance is m₂u₂ = m_in h_in with m₂ = m_in, giving c_v T₂ = c_p T_in, so T₂ = γ T_in. With T_in = 300 K and γ = 1.4, T₂ = 420 K: the gas is hotter than the line because the flow work p v pushed into the tank becomes internal energy.

⚠️ Units in the kinetic term
V²/2 is in m²/s² = J/kg. Adding it to h in kJ/kg without dividing by 1000 gives an error of a factor of 1000. Also check the sign of Ẇ and Q̇ against the stated convention before substituting.

Key takeaways

  • With Q positive into the system and W positive by the system, the closed-system first law is Q − W = ΔU + ΔKE + ΔPE; over a cycle ΣQ = ΣW.
  • For an ideal gas u and h depend only on T, so ΔU = m c_v ΔT and ΔH = m c_p ΔT for every process; c_p − c_v = R, c_v = R/(γ − 1), c_p = γR/(γ − 1).
  • Polytropic pVⁿ = C: W = mR(T₁ − T₂)/(n − 1) and Q = W(γ − n)/(γ − 1); an expanding gas with 1 < n < γ gains heat while cooling.
  • The steady-flow energy equation is Q̇ − Ẇ = ṁ[Δh + ΔV²/2 + gΔz]; enthalpy appears because it includes the flow work pv.
  • Standard devices: nozzle V₂² = V₁² + 2Δh drop; turbine and compressor w from Δh; throttle h₂ = h₁; mixing Σṁh in = Σṁh out. Divide V²/2 by 1000 for kJ/kg.
  • An initially evacuated adiabatic rigid tank filled from an ideal-gas line at T_in ends at T = γ T_in, since the flow work becomes internal energy.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. For an ideal gas, the internal energy depends on

    1. temperature only, whatever the pressure and volume
    2. temperature and pressure together
    3. volume only, because molecules do not interact
    4. pressure only, as it measures molecular momentum transfer
    Show answer

    Answer: A — temperature only, whatever the pressure and volume

    With no intermolecular forces, the energy of an ideal gas is purely the kinetic energy of its molecules, which is fixed by T. This is why du = c_v dT holds for every process, and why a free expansion of an ideal gas leaves T unchanged.

  2. An ideal gas has R = 287 J/kg·K and γ = c_p/c_v = 1.4. Its specific heat at constant pressure, in J/kg·K, is

    1. 717.5
    2. 287
    3. 410.0
    4. 1004.5
    Show answer

    Answer: D — 1004.5

    c_v = R/(γ − 1) = 287/0.4 = 717.5 and c_p = c_v + R = 1004.5 J/kg·K, which equals γR/(γ − 1) = 1.4 × 287/0.4. The value 717.5 is c_v, and 287 is the difference c_p − c_v.

  3. Use the convention that Q is positive into the system and W is positive for work done by the system. A closed gas of negligible kinetic and potential energy receives 200 kJ of heat while doing 120 kJ of work on its surroundings. Its internal energy changes by

    1. +320 kJ
    2. +80 kJ
    3. −80 kJ
    4. −320 kJ
    Show answer

    Answer: B — +80 kJ

    ΔU = Q − W = 200 − 120 = +80 kJ. Adding the magnitudes (+320 kJ) treats the work as done on the gas, which would be the opposite convention.

  4. A heat engine cycle (a closed system) receives heat transfers of 900 kJ and 300 kJ and rejects heat transfers of 400 kJ and 500 kJ during its four processes. The net work of the cycle is

    1. 1200 kJ
    2. 900 kJ
    3. 300 kJ
    4. 600 kJ
    Show answer

    Answer: C — 300 kJ

    Over a cycle ΔU = 0, so W_net = ΣQ = 900 + 300 − 400 − 500 = 300 kJ. The value 1200 kJ is the total heat received, 900 kJ is the total heat rejected, and 600 kJ comes from subtracting the 300 kJ input from the 900 kJ input.

  5. An ideal gas (γ = 1.4) expands polytropically with 1 < n < 1.4 from a higher to a lower temperature. Which statement is correct?

    1. The gas rejects heat, since its temperature falls
    2. The process is adiabatic, because the temperature changes
    3. The heat transfer is zero only if the process is also isothermal
    4. The gas receives heat, because the work done exceeds the fall in its internal energy
    Show answer

    Answer: D — The gas receives heat, because the work done exceeds the fall in its internal energy

    Q = W (γ − n)/(γ − 1) is positive when W > 0 and n < γ. Here W = −ΔU + Q exceeds −ΔU, so heat flows in even though T falls. A falling temperature does not imply heat rejection; Q = 0 would need n = γ.

  6. An adiabatic diffuser slows a steady gas stream from 200 m/s to 50 m/s with no shaft work and negligible change of elevation. Compared with the inlet, the exit specific enthalpy is

    1. higher, by (V₁² − V₂²)/2 per unit mass
    2. lower, since the stream loses kinetic energy
    3. equal, because no work or heat crosses the boundary
    4. higher, by the work done on the gas by the diffuser walls
    Show answer

    Answer: A — higher, by (V₁² − V₂²)/2 per unit mass

    The steady-flow energy equation with q = w = 0 gives h₂ = h₁ + (V₁² − V₂²)/2, so enthalpy rises by the kinetic energy lost. Equal enthalpy holds for a throttle, not a diffuser, since here the kinetic energy changes appreciably. The walls do no work on the gas.

  7. A real gas (not assumed ideal) flows steadily through an adiabatic throttling valve, with no shaft work and negligible changes of kinetic and potential energy. Which property has the same value upstream and downstream?

    1. Pressure, since the valve only restricts the flow
    2. Specific volume, since the mass flow rate is constant
    3. Specific enthalpy, from the steady-flow energy equation
    4. Specific internal energy, since no heat or work crosses the boundary
    Show answer

    Answer: C — Specific enthalpy, from the steady-flow energy equation

    With q = w = 0 and negligible ΔV²/2 and Δz, the steady-flow energy equation gives h₂ = h₁. The pressure falls and the specific volume rises. Internal energy is not conserved, because the flow work pv changes by (p₂v₂ − p₁v₁), so u₂ − u₁ equals the negative of that change for a real gas.

  8. Two streams of the same ideal gas at the same pressure, with constant c_p, enter an adiabatic mixing chamber that does no work: 2 kg/s at 300 K and 1 kg/s at 450 K. The mixed stream leaves at

    1. 375 K
    2. 350 K
    3. 400 K
    4. 337.5 K
    Show answer

    Answer: B — 350 K

    ṁ₁h₁ + ṁ₂h₂ = (ṁ₁ + ṁ₂)h₃ with h = c_pT gives T₃ = (2 × 300 + 1 × 450)/3 = 1050/3 = 350 K, the mass-flow-weighted mean. The simple average 375 K ignores the unequal flows.

  9. In the steady-flow energy equation the enthalpy h = u + pv appears, rather than the internal energy u, because

    1. the fluid always flows at constant pressure
    2. every flow process is isentropic
    3. pushing each unit mass across the boundary needs flow work pv
    4. the kinetic energy of a flowing fluid is defined to include its pressure
    Show answer

    Answer: C — pushing each unit mass across the boundary needs flow work pv

    Fluid entering the control volume has to be pushed in against the pressure there, and fluid leaving does work on the fluid ahead of it. This flow work pv per unit mass combines with u to give h. The pressure need not be constant.

  10. Which of the following statements about the specific heats of an ideal gas are correct?

    1. c_p − c_v = R, with R the specific gas constant.
    2. c_p is greater than c_v because a constant-pressure heating also does expansion work.
    3. c_v depends on the pressure at which the gas is held.
    4. The ratio γ = c_p/c_v is exactly 1.4 for every ideal gas.
    Show answer

    Answer: A — c_p − c_v = R, with R the specific gas constant.; B — c_p is greater than c_v because a constant-pressure heating also does expansion work.

    Mayer’s relation c_p − c_v = R follows from h = u + RT, and the extra c_p is the expansion work done per degree. For an ideal gas c_v depends on temperature only, and γ differs between gases (about 1.67 for monatomic ones).

  11. Use the convention that Q is positive into the system and W is positive by the system. Which of the following statements about a closed system are correct?

    1. The change Q − W between two states is the same for every path joining them.
    2. Q and W taken separately may depend on the path between the two states.
    3. For an isothermal process of an ideal gas, Q equals W.
    4. For an adiabatic process of an ideal gas, W is zero.
    Show answer

    Answer: A — The change Q − W between two states is the same for every path joining them.; B — Q and W taken separately may depend on the path between the two states.; C — For an isothermal process of an ideal gas, Q equals W.

    Q − W = ΔE is path-independent although Q and W are not; for an ideal gas ΔU = 0 in an isothermal process, so Q = W. In an adiabatic process Q = 0, not W: the work equals −ΔU.

  12. Which of the following statements about steady-flow devices are correct (kinetic energy changes in the turbine are neglected)?

    1. For an adiabatic turbine the specific work equals h₁ − h₂.
    2. For an adiabatic throttle with negligible kinetic energy change, the exit and inlet enthalpies are equal.
    3. An adiabatic nozzle delivers shaft work equal to its enthalpy drop.
    4. In an insulated heat exchanger the enthalpy lost by the hot stream equals that gained by the cold stream.
    Show answer

    Answer: A — For an adiabatic turbine the specific work equals h₁ − h₂.; B — For an adiabatic throttle with negligible kinetic energy change, the exit and inlet enthalpies are equal.; D — In an insulated heat exchanger the enthalpy lost by the hot stream equals that gained by the cold stream.

    The steady-flow energy equation gives w = h₁ − h₂ for the adiabatic turbine, h₂ = h₁ for the throttle, and ṁ_hΔh_h = −ṁ_cΔh_c for the heat exchanger. A nozzle has no shaft; its enthalpy drop becomes kinetic energy of the stream.

  13. A closed piston–cylinder contains 0.5 kg of an ideal gas with R = 0.287 kJ/kg·K and γ = 1.4. It is heated at constant pressure from 300 K to 450 K. Take Q positive into the gas. The heat added, in kJ, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 75.3

    c_p = γR/(γ − 1) = 1.4 × 0.287/0.4 = 1.0045 kJ/kg·K. Q = m c_p ΔT = 0.5 × 1.0045 × 150 = 75.34 kJ ≈ 75.3 kJ. Check: ΔU = 0.5 × 0.7175 × 150 = 53.81 kJ and W = 0.5 × 0.287 × 150 = 21.53 kJ sum to 75.34 kJ. Using c_v would give 53.8 kJ.

  14. Take air as an ideal gas with R = 0.287 kJ/kg·K and γ = 1.4. A mass of 2 kg expands polytropically (pVⁿ = constant) with n = 1.25, its temperature falling from 600 K to 500 K. Taking Q positive into the gas, the heat transfer in kJ, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 86.1

    W = mR(T₁ − T₂)/(n − 1) = 2 × 0.287 × 100/0.25 = 229.6 kJ. ΔU = m c_v ΔT = 2 × 0.7175 × (−100) = −143.5 kJ. Q = ΔU + W = 86.1 kJ. Check: W(γ − n)/(γ − 1) = 229.6 × 0.15/0.4 = 86.1 kJ. Taking the process as adiabatic would give Q = 0.

  15. Air (c_p = 1.005 kJ/kg·K, ideal gas) enters an adiabatic nozzle at 600 K with a velocity of 50 m/s and leaves at 500 K. There is no shaft work and the elevation change is negligible. The exit velocity, in m/s, to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 451

    The steady-flow energy equation gives V₂²/2 = V₁²/2 + c_p(T₁ − T₂). In J/kg: V₂² = 50² + 2 × 1005 × 100 = 2500 + 201 000 = 203 500 m²/s², so V₂ = 451.1 m/s ≈ 451 m/s. Forgetting the factor 1000 between kJ and J would give a tiny velocity.

  16. Steam flows steadily at 5 kg/s through a turbine. At inlet h₁ = 3230 kJ/kg and V₁ = 50 m/s; at exit h₂ = 2550 kJ/kg and V₂ = 150 m/s. The turbine loses heat to the surroundings at a rate of 40 kW. Neglect elevation changes. The power output of the turbine, in kW, to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 3310

    With Q̇ = −40 kW taken positive into the fluid: Ẇ = Q̇ − ṁ[(h₂ − h₁) + (V₂² − V₁²)/2000] = −40 − 5 × [−680 + (22 500 − 2500)/2000] = −40 − 5 × (−680 + 10) = −40 + 3350 = 3310 kW. Ignoring the heat loss gives 3350 kW, and ignoring kinetic energy gives 3360 kW.

  17. Air flows steadily at 2 kg/s through a compressor and is heated by compression from 300 K to 440 K. Take air as an ideal gas with c_p = 1.005 kJ/kg·K, neglect kinetic and potential energy changes, and let the compressor lose heat to the surroundings at 30 kW. The shaft power input, in kW, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 311.4

    Ẇ = Q̇ − ṁ c_p ΔT = −30 − 2 × 1.005 × 140 = −30 − 281.4 = −311.4 kW, so the input is 311.4 kW. The heat loss adds to the work required; subtracting it would give 251.4 kW.

  18. A rigid, adiabatic tank is initially evacuated and is connected through a valve to a line carrying an ideal gas at a constant 300 K, with γ = 1.4 and constant specific heats. The valve is opened until flow stops. Neglect kinetic and potential energies. The final temperature of the gas in the tank, in K, to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 420

    The energy balance for the tank is m₂u₂ = m_in h_in with m₂ = m_in, so c_v T₂ = c_p T_in and T₂ = γ T_in = 1.4 × 300 = 420 K. Expecting T₂ = 300 K would ignore the flow work done on the gas entering the tank.