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Thermodynamics · Chapter 2 of 10 · Beginner

Energy, Heat, and Work

Energy lives inside a system as internal, kinetic, and potential energy, and it crosses the boundary in just two ways: heat and work. The energy balance that tracks all of this is the first law.

01

Readiness check

This chapter builds the energy balance. Tick only what you can do closed-notes.

  • Write kinetic energy ½mV² and potential energy mgz with units.
  • Recall power as energy per unit time, in watts.
  • Use Q = mcΔT for sensible heating.
  • Name a system boundary and what crosses it.
  • Convert kJ, kW, and minutes consistently.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsRevisit systems and state in Chapter 1.
3 or more weak itemsReview work and energy in Physics: First Law.
02

The core idea

The total energy of a system can change only by heat or work crossing its boundary. That bookkeeping statement is the first law: the energy in minus the energy out equals the change stored.

E = U + KE + PEEin − Eout = ΔEsystemQ − W = ΔE

Energy inside the system is internal (U, the microscopic thermal and molecular energy) plus the bulk kinetic and potential energy. Energy crosses the boundary as heat Q, driven by a temperature difference, or as work W, every other mode: boundary, shaft, electrical. We take heat into the system and work done by the system as positive. The first law simply says energy is conserved once you count every crossing.

The skill works when: every energy stream is identified, given a sign, and matched to a change in stored energy.
The skill breaks down when: heat and work are double counted, or an energy stream is missed entirely.
The concept. Stored energy E sits inside the boundary. Heat enters and work leaves; the net of the two equals the change in stored energy. That is the first law as an energy balance.
03

The skills, taught in order

Five skills name the forms of energy, the two transfer modes, the idea of efficiency, and the energy balance that ties them together.

2.1 Forms of energy

The total energy of a system is E = U + KE + PE, where internal energy U is the microscopic energy of the molecules, kinetic energy is ½mV², and potential energy is mgz. Per unit mass, e = u + V²/2 + gz. In most closed-system problems the bulk kinetic and potential terms do not change, so only the internal energy matters.

2.2 Heat transfer

Heat is energy that crosses a boundary because of a temperature difference, always from hot to cold. It is a path quantity, not a property: a system has no stored heat. A process with no heat transfer is adiabatic, achieved by insulation or by happening too quickly to exchange heat.

2.3 Work transfer

Work is any energy crossing the boundary that is not heat. Common modes are boundary (a moving piston, W = ∫P dV), shaft (a rotating turbine or pump), and electrical. Like heat, work is a path quantity. Power is the rate of doing work, in watts.

2.4 Mechanical energy and efficiency

The mechanical energy of a flowing fluid is emech = Pv + V²/2 + gz, the part that a pump or turbine can convert directly to work. Real devices fall short: efficiency is useful output over required input, η = Eout/Ein. A turbine-generator set multiplies its component efficiencies.

Energy streamSymbolSign convention
Heat into the systemQpositive
Heat out of the systemQnegative
Work done by the systemWpositive
Work done on the systemWnegative

2.5 The energy balance

For any system, Ein − Eout = ΔEsystem. For a closed system this is Q − W = ΔE, and when only internal energy changes, Q − W = ΔU. In rate form, the same balance reads Q̇ − Ẇ = dE/dt, which is how steady devices are analysed.

Engineering connection: the same balance sizes a water heater, a pump motor, and a power-plant turbine; only the streams that matter change from one to the next.

04

Worked example 1: power from a hydro reservoir

Water flows from a reservoir 85 m above a turbine at 0.50 m³/s. With water density 1000 kg/m³ and an overall turbine-generator efficiency of 84 percent, find the ideal power available and the electrical power produced.

Figure 1. The 85 m drop sets the mechanical energy per kilogram (gz). The mass flow turns it into ideal power; efficiency reduces it to the electrical output.
  1. ProblemFind the ideal and electrical power for the hydro arrangement in Figure 1.
  2. Given / findz = 85 m, Q = 0.50 m³/s, ρ = 1000 kg/m³, η = 0.84, g = 9.81 m/s². Find ideal power and electrical power.
  3. AssumptionsSteady flow; the only mechanical energy is the elevation term; velocity and pressure terms at the surfaces are negligible.
  4. ModelMass flow ṁ = ρQ; ideal power is ṁgz; the actual electrical power is η times that.
  5. Equationsṁ = ρQẆideal = ṁgzẆelec = η Ẇideal
  6. Solveṁ = 1000 × 0.50 = 500 kg/s. Ẇideal = 500 × 9.81 × 85 = 416 925 W = 417 kW. Ẇelec = 0.84 × 417 = 350 kW.
  7. CheckUnits: (kg/s)(m/s²)(m) = kg·m²/s³ = W. The mechanical energy per kilogram is gz = 834 J/kg, a small number that only becomes large power because the flow is 500 kg/s.
  8. ConclusionMechanical energy times mass flow gives power, and efficiency converts the ideal value to the usable one. The same logic runs every pump and turbine in Chapter 10.
Result. Ideal power 417 kW; with 84 percent efficiency, 350 kW of electricity.
05

Worked example 2: heating water with an electric element

A 4 kW electric element heats 50 kg of water from 15 °C to 45 °C in a well-insulated tank. With water specific heat 4.18 kJ/kg·K, find the heat required and the time it takes.

Figure 2. The insulated tank is a closed system with electrical work in and no heat loss, so all the electrical energy goes into raising the water temperature.
  1. ProblemFind the energy and time to heat the water in Figure 2.
  2. Given / findP = 4 kW, m = 50 kg, ΔT = 30 K, c = 4.18 kJ/kg·K. Find Q and the time t.
  3. AssumptionsNo heat loss (insulated); water properties constant; all electrical work becomes internal energy.
  4. ModelEnergy balance Q = mcΔT for the sensible heating; time follows from t = Q/P.
  5. EquationsQ = mcΔTt = Q/Ẇ
  6. SolveQ = 50 × 4.18 × 30 = 6270 kJ. Then t = 6270/4 = 1568 s = 26.1 min.
  7. CheckUnits: kJ divided by kW (kJ/s) gives seconds. A few kilowatt-hours to heat a tank of water is the right order of magnitude for a domestic heater.
  8. ConclusionThe energy balance turns a power rating into a heating time. Any real heat loss would lengthen the time, which is why insulation matters.
Result. Q = 6270 kJ, and the 4 kW element takes about 26 minutes.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Treating heat as storedTalking about the heat contained in a body"Is this crossing a boundary, or sitting inside?"Stored energy is internal energy; heat is only the transfer.
Sign-convention driftΔE comes out with the wrong sign"Is heat in positive and work out positive in my balance?"Fix one convention and label every stream on the diagram.
Dropping efficiencyPredicted output exceeds the input"Did I multiply by η somewhere?"Useful output is always less than input; apply η once.
Mismatched unitsTime or power off by factors of 1000 or 60"Are kJ, kW, and seconds consistent?"Keep energy in kJ, power in kW, and convert time deliberately.
07

Practice ladder

Level 1 · Direct skill

A 2 kW kettle heats 1.5 kg of water from 20 °C to 100 °C. With c = 4.18 kJ/kg·K, how long does it take (ignoring losses)?

Show answer

Q = mcΔT = 1.5 × 4.18 × 80 = 501.6 kJ. Time t = Q/P = 501.6/2 = 250.8 s ≈ 4.2 min. Real kettles take a little longer because of heat loss.

Level 2 · Mixed concept

The hydro plant (Worked Example 1) keeps the same head but the flow doubles to 1.0 m³/s. What is the new electrical power, and why?

Show answer

Power scales linearly with mass flow, so it doubles to about 700 kW. The mechanical energy per kilogram (gz = 834 J/kg) is unchanged; only the kilograms per second doubled.

Level 3 · Independent problem

A pump motor draws 5 kW and delivers water with a useful mechanical-energy rate of 4.1 kW. What is its efficiency, and where does the missing 0.9 kW go?

Show answer

η = 4.1/5 = 0.82, or 82 percent. The missing 0.9 kW becomes internal energy (heat) in the motor windings, bearings, and the water through friction.

Level 4 · Transfer to real engineering

Pick a device that converts energy (a fan, a blender, an e-bike). Estimate its input power and useful output, and state its efficiency and where the losses go.

What good work looks like

A named input and useful output with units, an efficiency below 100 percent, and a sensible account of the losses as heat.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my energy balance has every stream signed correctly."
"Give me four scenarios; I will say whether energy crosses as heat or work."
"How long does the heater take?" Doing Q = mcΔT and t = Q/P yourself is the skill.
"What is the efficiency?" Forming the output-over-input ratio is the point.

Portfolio task

Write a one-page energy balance for a real heating or power device: the system, the streams crossing as heat and work, a calculated result, and an efficiency statement.

Must include: a labelled boundary, signed energy streams, and an efficiency below one with the losses named.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Name the three forms of stored energy.

Internal energy U, kinetic energy ½mV², and potential energy mgz; together they make the total energy E.

2. How do heat and work differ?

Heat crosses the boundary because of a temperature difference; work is every other transfer. Both are path quantities, not properties.

3. State the energy balance for a closed system.

Q − W = ΔE, which reduces to Q − W = ΔU when only internal energy changes.

4. Define efficiency.

Useful energy output divided by required energy input, always less than one for a real device.

5. Write power for a fluid raised a height z.

Ẇ = ṁgz, with ṁ = ρQ the mass flow rate.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the hydro power and heating time from a blank page.
+3 daysWrite an energy balance for two new devices.
+7 daysCarry the balance into pure substances, Chapter 3.
+30 daysReuse the first law for closed and open systems in Chapters 4 and 5.
10

Textbook mapping

ItemMapping
Primary sourceBorgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 4 (Work and Heat) and Section 2.6 (Energy)
Cross-referenceÇengel and Boles, Ch. 2 · Moran, Ch. 2
Core topics2.1 Forms of energy · 2.2 Heat · 2.3 Work · 2.4 Mechanical energy and efficiency · 2.5 The energy balance
Engineering connectionThe energy balance sizes heaters, pumps, and turbines throughout the course.
Read nextChapter 3: Properties of Pure Substances.