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Thermodynamics

3% of exam

First and second laws, energy, heat, and work, efficiencies and coefficient of performance, conduction, convection, and radiation heat transfer, and the behavior of ideal gases.

3 concepts

A. Thermodynamic laws

Thermodynamic Laws, Efficiency, and COP

The first and second laws, the Carnot absolute-temperature ceiling, refrigeration and heat-pump COP, and the one-way direction of entropy.

Thermodynamics shows up on the FE Environmental wherever energy is converted or moved: a waste-to-energy steam cycle, the work to drive a refrigeration compressor, the heat recovered from a digester, the efficiency ceiling on any power plant whose cooling water you are permitting. The physics is rarely hard — these are unit-and-reference-frame problems. Two habits cost more points than anything else: feeding Celsius into a formula that demands absolute temperature, and treating a coefficient of performance as if it were an efficiency that cannot exceed one. Keep temperatures in kelvin, write down which energy you are calling 'useful,' and these become free points. (FE Reference Handbook 10.6 — Thermodynamics)

First law: energy is conserved

The first law is conservation of energy written for a control volume — energy in equals energy out plus the change in storage. For a heat engine over one complete cycle there is no net change in internal energy, so the heat taken in at the hot source equals the net work plus the heat rejected:

Q_H = W + Q_L

. For a flowing stream heated or cooled without phase change, the rate you almost always need is the sensible-heat balance, where

\dot{m}

is the mass flow rate (kg/s),

c_p

the specific heat (J/kg·K, about

4{,}186

for water and roughly

1{,}000

–

1{,}100

for combustion gases), and

\Delta T = T_{out}-T_{in}

. Because

\Delta T

is a difference it has the same numerical value in K or °C — that is the one place Celsius is safe.

\dot{Q} = \dot{m}\,c_p\,\Delta T \qquad\text{(cycle: } Q_H = W + Q_L\text{)}

Thermal efficiency of a heat engine

A heat engine takes in

Q_H

at high temperature, delivers net work

W

, and dumps the remainder

Q_L

to a cold sink. Its thermal efficiency is the fraction of the input heat that becomes useful work. The denominator is always the heat purchased (

Q_H

), never the work — a classic slip. The rest,

Q_L

, is not a design defect you can engineer away; the second law guarantees some heat must be rejected.

\eta = \frac{W}{Q_H} = \frac{Q_H - Q_L}{Q_H} = 1 - \frac{Q_L}{Q_H}

The second law and the Carnot ceiling

The second law sets a hard upper bound on

\eta

that no real machine can beat. For any engine operating between two reservoirs, the maximum possible efficiency — the Carnot efficiency — depends only on the two absolute temperatures, not on the working fluid or the cycle details. The two temperatures MUST be absolute (K or °R). This single relationship explains why power plants chase the highest practical boiler temperature and the lowest practical condenser temperature, and why a plant rejecting heat to a warm river in August loses efficiency. A real plant's actual efficiency divided by

\eta_{Carnot}

is its second-law (exergetic) efficiency — how close it comes to the thermodynamic ideal.

\eta_{Carnot} = 1 - \frac{T_L}{T_H}

Run it backward: refrigerators and heat pumps

Reverse the cycle and you spend work

W

to pump heat from cold to hot — a refrigerator, air conditioner, or heat pump. You rate it not by efficiency but by coefficient of performance: useful heat moved per unit of work. The 'useful' effect depends on the device. A refrigerator or air conditioner values the heat pulled out of the cold space (

Q_L

); a heat pump values the heat delivered to the warm space (

Q_H

). Because

Q_H = Q_L + W

, the two COPs differ by exactly one:

\mathrm{COP}_{HP} = \mathrm{COP}_{ref} + 1

. Both are routinely greater than 1 — that is the whole point of a heat pump, and it does NOT violate energy conservation because you are moving heat, not creating it.

\mathrm{COP}_{ref} = \frac{Q_L}{W}, \qquad \mathrm{COP}_{HP} = \frac{Q_H}{W}

The COP ceiling and entropy direction

Just as Carnot caps efficiency, a reversed-Carnot cycle caps the COP between the same two reservoirs:

\mathrm{COP}_{HP,\max} = T_H/(T_H-T_L)

and

\mathrm{COP}_{ref,\max} = T_L/(T_H-T_L)

, both with absolute temperatures. The deeper statement underneath all of this is the increase-of-entropy principle: for any real (irreversible) process the total entropy of system plus surroundings must increase, and only a reversible process holds it constant. Heat flows spontaneously from hot to cold, never the reverse, because the hot-to-cold direction is the one that raises total entropy. That arrow is what makes

Q_L

unavoidable and what every efficiency and COP ceiling ultimately encodes.

\Delta S_{total} = \Delta S_{system} + \Delta S_{surr} \ge 0

Exam strategy

Convert every temperature to kelvin the instant you read it — write

T_H

and

T_L

in K before touching a formula. Decide up front whether the device produces work (use

\eta

) or consumes work to move heat (use COP), and for COP name the useful stream (

Q_L

for cooling,

Q_H

for heating). Carnot

\eta

and Carnot COP are ceilings: a real device must come in below the Carnot value, so a computed actual efficiency above the Carnot number means you made an error (usually Celsius in the formula). Remember

1\ \text{ton refrigeration} = 12{,}000\ \text{Btu/hr} = 3{,}516\ \text{W}

, and keep

Q_H = W + Q_L

handy to recover whichever quantity is missing.

Key equations

First law for a cycle

Q_H = W + Q_L

Over one complete cycle internal energy returns to its start; heat in equals net work plus heat rejected. Energies in consistent units (J, kJ, or W for rates).

Sensible-heat rate (first law, flowing stream)

\dot{Q} = \dot{m}\,c_p\,\Delta T

$\dot{m}$ = mass flow (kg/s), $c_p$ = specific heat (J/kg·K), $\Delta T = T_{out}-T_{in}$

Thermal efficiency

\eta = \frac{W}{Q_H} = 1 - \frac{Q_L}{Q_H}

Carnot (maximum) efficiency

\eta_{Carnot} = 1 - \frac{T_L}{T_H}

Second-law (exergetic) efficiency

\eta_{II} = \frac{\eta_{actual}}{\eta_{Carnot}}

Refrigeration COP

\mathrm{COP}_{ref} = \frac{Q_L}{W}

Heat-pump COP

\mathrm{COP}_{HP} = \frac{Q_H}{W} = \mathrm{COP}_{ref} + 1

Carnot COP ceilings

\mathrm{COP}_{HP,\max} = \frac{T_H}{T_H-T_L}, \quad \mathrm{COP}_{ref,\max} = \frac{T_L}{T_H-T_L}

Increase-of-entropy principle

\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} \ge 0

Reservoir entropy change

\Delta S_{res} = \frac{Q}{T_{res}}

Worked examples

Carnot ceiling and second-law efficiency of a waste-to-energy plant

Problem. A waste-to-energy steam plant boils at $540^\circ\text{C}$ and condenses at $40^\circ\text{C}$ . (a) What is the maximum possible thermal efficiency? (b) The plant actually converts $27\%$ of the heat in the refuse to electricity — what is its second-law efficiency?

Solution. Convert to absolute temperature first:

T_H = 540 + 273.15 = 813.15\ \text{K}

,

T_L = 40 + 273.15 = 313.15\ \text{K}

\eta_{Carnot} = 1 - \frac{313.15}{813.15} = 0.615

Heat-pump work, absorbed heat, and the Carnot limit

Problem. A heat pump must deliver $12\ \text{kW}$ of heating to a building. Its rated $\mathrm{COP}_{HP} = 3.5$ . (a) Find the electrical work input and the heat drawn from the outside air. (b) If indoors is $21^\circ\text{C}$

First-law energy balance on cooling incinerator flue gas

Problem. Flue gas leaves an incinerator at $2.5\ \text{kg/s}$ and $320^\circ\text{C}$ and is cooled to $150^\circ\text{C}$ in a heat-recovery boiler. Take $c_p = 1.05\ \text{kJ/(kg·K)}$

Common pitfalls

•Using Celsius in Carnot or COP formulas. Any ratio of absolute temperatures requires kelvin (or rankine) — $T_L/T_H$ with °C is meaningless. Differences ( $\Delta T$ ) are the only place °C is safe.
•Dividing by $W$ instead of $Q_H$ in efficiency. $\eta = W/Q_H$ , not $W/Q_L$ or $Q_H/W$ . The denominator is the heat you paid for.
•Treating COP like an efficiency that must be below 1. COP routinely exceeds 1 (often 3–5) because you are moving heat, not converting it; a value below 1 usually signals an error.
•Confusing refrigerator and heat-pump COP. Cooling values $Q_L$ ; heating values $Q_H$ ; they differ by exactly 1 ( $\mathrm{COP}_{HP} = \mathrm{COP}_{ref}+1$
•Getting a real efficiency above the Carnot value. Impossible — the Carnot result is a ceiling. If your 'actual' beats it, recheck that you used absolute temperatures and the right $Q_H$ .
•Forgetting that heat must be rejected. $Q_L = 0$ would mean $100\%$ efficiency; the second law forbids it, so always account for the rejected heat in an energy balance.
•Mixing energy and power. $Q$ and $W$ in J (or kJ) for a per-cycle problem, but in W (J/s) for a rate problem — do not multiply a flow rate by a per-cycle energy.

References

NCEES FE Reference Handbook — Thermodynamics
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics — first/second law, Carnot, refrigeration cycles
Davis & Cornwell, Introduction to Environmental Engineering — energy recovery and waste-to-energy context

B. Energy, heat, and work

Heat Transfer: Conduction, Convection, and Radiation

Fourier conduction, thermal resistance in series, Newton convection and the overall U, composite-wall networks, and the Stefan-Boltzmann radiation law.

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C. Behavior of ideal gases

Ideal-Gas Behavior

The ideal-gas law, universal vs specific gas constant, molar volume, Dalton partial pressures, mixture density, and the ppm-to-mg/m3 conversion.

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(K or °C, same magnitude). Gives rate in W.

Fraction of input heat converted to net work. Denominator is the heat supplied $Q_H$ , never $W$ .

Upper bound for any engine between reservoirs at absolute temperatures $T_H$ and $T_L$ (K or °R). No real engine can exceed it.

How close a real device comes to the reversible ideal; dimensionless, always < 1 for real machines.

Useful cooling (heat removed from cold space $Q_L$ ) per unit work input. For refrigerators and air conditioners.

Useful heating (heat delivered to warm space $Q_H$ ) per unit work input.

Maximum COP from a reversed-Carnot cycle; absolute temperatures only.

Total entropy never decreases; equality only for a reversible process. Fixes the hot-to-cold direction of spontaneous heat flow.

Entropy change of a thermal reservoir exchanging heat $Q$ at constant absolute temperature $T_{res}$ (J/K).

. (a) Carnot efficiency:

\eta_{Carnot} = 1 - T_L/T_H = 1 - 313.15/813.15 = 1 - 0.3851 = 0.615

, i.e.

61.5\%

. (b) Second-law efficiency:

\eta_{II} = \eta_{actual}/\eta_{Carnot} = 0.27/0.615 = 0.439

, i.e.

43.9\%

. Final:

\eta_{Carnot} = 61.5\%

;

\eta_{II} = 43.9\%

. Sanity check: the actual

27\%

is comfortably below the

61.5\%

ceiling, as the second law requires, and the second-law efficiency lands in the typical

40

–

50\%

band for a steam Rankine plant — both consistent.

and outdoors is

0^\circ\text{C}

, what is the lowest work input thermodynamically possible?

Solution. (a)

W = Q_H/\mathrm{COP}_{HP} = 12/3.5 = 3.43\ \text{kW}

. By the first law,

Q_L = Q_H - W = 12 - 3.43 = 8.57\ \text{kW}

pulled from the outside air. (b) Carnot ceiling with

T_H = 294.15\ \text{K}

,

T_L = 273.15\ \text{K}

:

\mathrm{COP}_{HP,\max} = T_H/(T_H-T_L) = 294.15/21.0 = 14.0

. The minimum work is

W_{\min} = Q_H/\mathrm{COP}_{HP,\max} = 12/14.0 = 0.857\ \text{kW}

. Final:

W = 3.43\ \text{kW}

,

Q_L = 8.57\ \text{kW}

,

W_{\min} = 0.857\ \text{kW}

. Sanity check: the real

3.43\ \text{kW}

exceeds the

0.857\ \text{kW}

ideal (real

>

ideal work, real COP

<

Carnot COP), exactly as the second law demands.

W = \frac{Q_H}{\mathrm{COP}_{HP}} = \frac{12}{3.5} = 3.43\ \text{kW}

. What rate of heat is recovered?

Solution. Sensible-heat balance:

\dot{Q} = \dot{m}\,c_p\,\Delta T

. Here

\Delta T = 320 - 150 = 170\ \text{K}

(the magnitude is identical in K and °C).

\dot{Q} = 2.5 \times 1.05 \times 170 = 446\ \text{kJ/s} = 446\ \text{kW}

. Final:

\dot{Q} = 446\ \text{kW}

recovered. Sanity check: units are

(\text{kg/s})(\text{kJ/kg·K})(\text{K}) = \text{kJ/s} = \text{kW}

; magnitude (a few hundred kW from a multi-kg/s hot stream) is reasonable for a small recovery boiler.

\dot{Q} = 2.5\,(1.05)(170) = 446\ \text{kW}

). Pick the one matching the useful effect.