First Law of Thermodynamics
5. Energy Conservation Applications
Learning outcomes
- I can analyze thermodynamic systems using the First Law.
- I can solve energy balance problems.
- I can explain energy transfer in engines and compressors.
- I can identify assumptions used in thermodynamic calculations.
- I can apply conservation of energy to engineering systems.
Energy Conservation Applications
The First Law of Thermodynamics allows us to analyse real systems by tracking how energy enters, leaves, and is stored.
For a thermodynamic system, energy may:
- enter or leave by heating
- enter or leave through work
- change the internal energy of the system
Using the sign convention from the previous notes:
ΔU = Q − W
where:
- ΔU = change in internal energy
- Q = energy transferred to the system by heating
- W = work done by the system on its surroundings
The First Law is fundamentally an energy balance. Energy cannot simply appear or disappear.
Thinking in Terms of a System
Before solving an energy-conservation problem, we must decide what we are studying.
This is called the system.
Everything outside the system is called the surroundings.
For example, if we study the gas inside a piston-cylinder:
System: gas
Surroundings: piston, cylinder and everything outside the gas
Energy can cross the system boundary.
Once the system has been identified, ask:
- What energy enters?
- What energy leaves?
- Is work done by or on the system?
- Does the internal energy change?
This approach is often more useful than immediately searching for an equation.
Energy Balance
A simple energy balance can be thought of as:
energy entering − energy leaving = increase in stored energy
For the simple thermodynamic systems considered here:
ΔU = Q − W
Suppose 2,500 J enters a gas by heating and the gas does 900 J of work.
Then:
ΔU = 2500 − 900
ΔU = +1600 J
The energy accounting is:
2500 J in = 1600 J stored + 900 J out
Nothing has been created or destroyed.
Energy-Flow Diagrams
An energy-flow diagram can make an energy balance easier to understand.
Suppose:
1800 J heating → [ SYSTEM ] → 700 J work
The change in internal energy is:
ΔU = 1800 − 700
ΔU = +1100 J
The system gains 1,100 J of internal energy.
If instead:
600 J heating → [ SYSTEM ] → 1000 J work
then:
ΔU = 600 − 1000
ΔU = −400 J
The system loses 400 J of internal energy.
A Systematic Problem-Solving Method
For an energy-balance problem, use the following process:
- Define the system.
- Identify all known energy transfers.
- Determine which transfers enter and which leave.
- Choose and maintain a consistent sign convention.
- Write the First Law.
- Substitute the known values.
- Solve for the unknown quantity.
- Interpret the sign and units of the answer.
- Check whether the answer is physically reasonable.
This method works for many thermodynamic systems.
Worked Example: Simple Energy Balance
A gas receives 3,200 J of energy by heating and does 1,400 J of work on its surroundings.
Calculate its change in internal energy.
Use:
ΔU = Q − W
Substitute:
ΔU = 3200 − 1400
ΔU = +1800 J
The positive result means the internal energy increases by 1,800 J.
Finding an Unknown Heat Transfer
A gas does 600 J of work while its internal energy increases by 900 J.
Find Q.
Start with:
ΔU = Q − W
Rearrange:
Q = ΔU + W
Substitute:
Q = 900 + 600
Q = 1500 J
Therefore, 1,500 J of energy was transferred into the gas by heating.
Finding Unknown Work
A gas receives 4,000 J by heating while its internal energy increases by 2,500 J.
Find the work done by the gas.
Start with:
ΔU = Q − W
Rearrange:
W = Q − ΔU
Substitute:
W = 4000 − 2500
W = 1500 J
The gas does 1,500 J of work on its surroundings.
Engines
A heat engine is a device that converts part of an energy transfer from a high-temperature source into useful mechanical work.
Examples include:
- internal combustion engines
- steam engines
- gas turbines
- steam turbines
A simplified engine involves:
high-temperature source → working substance → useful work + energy transferred to cooler surroundings
Not all of the energy entering an engine becomes useful work.
Energy Balance in an Engine
Suppose an engine receives:
Q_in = 5000 J
and transfers:
Q_out = 3200 J
to the cooler surroundings.
If the engine completes a cycle, it returns to its original thermodynamic state.
Therefore:
ΔU = 0
The energy balance becomes:
energy in = useful work + energy out
Therefore:
W = 5000 − 3200
W = 1800 J
The engine produces 1,800 J of mechanical work per cycle.
Why Engines Cannot Simply Convert All Input Energy to Work
Real heat engines must transfer some energy to their surroundings.
They also experience effects such as:
- friction
- heating of engine components
- exhaust losses
- sound
- incomplete combustion
- other irreversible processes
Consequently, only part of the supplied energy becomes useful mechanical work.
Engine Efficiency
The efficiency of an engine compares useful energy output with energy input.
efficiency = useful energy output / total energy input
As a percentage:
efficiency = (useful energy output / total energy input) × 100%
Example
An engine receives 8,000 J and produces 2,800 J of useful work.
Efficiency:
efficiency = 2800 / 8000 × 100%
efficiency = 35%
Therefore, 35% of the input energy becomes useful work.
The remaining energy is transferred elsewhere.
Worked Engine Problem
A heat engine receives 12,000 J from a high-temperature source during each cycle and transfers 7,500 J to its surroundings.
Calculate the useful work.
W = Q_in − Q_out
W = 12000 − 7500
W = 4500 J
Now calculate the efficiency:
efficiency = 4500 / 12000 × 100%
efficiency = 37.5%
So the engine produces 4,500 J of work per cycle with an efficiency of 37.5%.
Compressors
A compressor uses mechanical work to increase the pressure of a gas.
Instead of obtaining work from an expanding gas, we do work on the gas.
Examples include compressors in:
- refrigerators
- air conditioners
- compressed-air systems
- gas pipelines
- industrial equipment
- heat pumps
Energy Transfer in a Compressor
When a gas is compressed:
- work is done on the gas
- energy enters the gas
- its internal energy may increase
- its temperature may increase
Using our convention, W represents work done by the system.
Therefore, when work is done on the gas:
W is negative
Example
800 J of work is done on a gas, and no energy is transferred by heating.
Therefore:
Q = 0
W = −800 J
Using:
ΔU = Q − W
ΔU = 0 − (−800)
ΔU = +800 J
The gas gains 800 J of internal energy.
Compressor with Energy Loss
Suppose 2,000 J of work is done on a gas while 500 J of energy leaves the gas by heating.
Using our sign convention:
Q = −500 J
W = −2000 J
Therefore:
ΔU = Q − W
ΔU = −500 − (−2000)
ΔU = +1500 J
The gas's internal energy increases by 1,500 J.
The energy accounting can also be viewed more simply:
2000 J in by work − 500 J out by heating = 1500 J stored
Both approaches give the same result.
Engines and Compressors Compared
| Engine | Compressor |
|---|---|
| Gas often expands | Gas is compressed |
| Gas does work | Work is done on gas |
| Energy transferred out as useful work | Mechanical energy transferred into gas |
| Converts thermal energy transfer into mechanical output | Uses mechanical input to change gas state |
| Common in cars and turbines | Common in refrigeration and air systems |
The First Law applies to both. The main difference is the direction of energy transfer.
Engineering Energy Systems
The First Law can be applied to much more than a sealed gas cylinder.
Engineers use energy balances to analyse systems such as:
- engines
- turbines
- compressors
- pumps
- boilers
- heat exchangers
- refrigerators
- power stations
- air-conditioning systems
The details can become more complex, but the basic question remains:
Where does the energy come from, where does it go, and how much is stored?
Turbines
A turbine extracts energy from a moving or expanding fluid.
For example, high-temperature steam can enter a turbine and expand.
As it passes through the turbine:
- the steam transfers energy
- turbine blades rotate
- mechanical work is produced
- the steam leaves with less available energy
Power Stations
A simplified thermal power station involves several energy transfers:
fuel or other energy source → heating of working fluid → expanding steam or gas → turbine → generator
At each stage, conservation of energy applies.
For example, if a turbine receives an effective energy input of 1,000 MJ and produces 420 MJ of useful mechanical output, the remaining energy has not disappeared.
It must be accounted for through:
- energy remaining in the working fluid
- energy transferred to cooling systems
- friction
- sound
- other losses to the surroundings
Energy conservation applies even when a machine is inefficient.
Refrigerators and Heat Pumps
A refrigerator does not "create cold."
Instead, it uses work to transfer energy from a cooler region to a warmer region.
A simplified energy balance is:
energy removed from refrigerator + electrical work input = energy transferred to room
For example, suppose a refrigerator removes 400 J from its interior while receiving 150 J of electrical work.
The energy transferred to the room is:
400 + 150 = 550 J
This explains why the back of a refrigerator releases more energy to the room than is removed from the refrigerated compartment.
Assumptions in Thermodynamic Calculations
Real engineering systems are complicated. To make calculations manageable, we often make assumptions.
An assumption simplifies the system while keeping the model useful.
Common assumptions include the following.
Negligible Heat Loss
We may assume:
Q ≈ 0
This is approximately reasonable when a process occurs rapidly or the system is well insulated.
Such a process may be treated as adiabatic.
Constant Pressure
For some expansion problems, we assume the external pressure remains constant.
This allows:
W = pΔV
Ideal Gas Behaviour
We may assume the gas behaves as an ideal gas.
This neglects molecular volume and intermolecular forces under conditions where these effects are relatively unimportant.
Negligible Friction
We may ignore friction in an idealized engine or piston.
In a real machine, friction transfers energy into internal energy and reduces useful mechanical output.
No Change in Kinetic Energy
If the system as a whole is not accelerating significantly, changes in macroscopic kinetic energy may be ignored.
No Change in Gravitational Potential Energy
If height changes are small, gravitational potential-energy changes may be treated as negligible.
Why Assumptions Matter
Every mathematical model is based on assumptions.
Consider a piston problem where we calculate:
W = pΔV
This assumes that the relevant pressure can be treated appropriately over the expansion.
If pressure changes significantly, simply using one pressure value may not accurately represent the work.
Similarly, assuming no heat loss in an engine may make a classroom calculation easier, but real engines always transfer some energy to their surroundings.
A good solution should therefore state important assumptions when they affect the answer.
Ideal vs. Real Systems
An ideal system is a simplified model.
A real system includes additional effects.
For example:
Ideal Compressor
We might assume:
- no energy loss by heating
- no friction
- perfect insulation
Real Compressor
In reality:
- components become warm
- friction occurs
- sound is produced
- energy transfers to the surroundings
The First Law still holds in both cases.
The difference is that the real system requires more energy transfers to be included in the balance.
Worked Engineering Problem
A compressor receives 5,000 J of mechanical work. During compression, 1,200 J of energy is transferred from the gas to the surroundings by heating.
Find the change in internal energy.
Energy entering by work:
5000 J
Energy leaving by heating:
1200 J
Therefore:
ΔU = 5000 − 1200
ΔU = +3800 J
The internal energy of the gas increases by 3,800 J.
Using the formal sign convention gives the same result:
Q = −1200 J
W = −5000 J
ΔU = Q − W
ΔU = −1200 − (−5000)
ΔU = +3800 J
Worked Engineering Problem: Engine
An engine receives 20,000 J of energy from combustion during one cycle. It produces 6,000 J of useful mechanical work and returns to its initial state.
Because it completes a cycle:
ΔU = 0
Therefore, the remaining energy must leave the system:
Q_out = 20000 − 6000
Q_out = 14000 J
The engine transfers 14,000 J to the surroundings.
Its efficiency is:
efficiency = 6000 / 20000 × 100%
efficiency = 30%
Multi-Step Energy Balance
Sometimes several transfers occur.
Suppose a system:
- receives 4,000 J by heating
- receives 1,500 J through work done on it
- loses 800 J by heating
Instead of worrying about signs immediately, we can first organize the transfers.
Total energy entering:
4000 + 1500 = 5500 J
Energy leaving:
800 J
Net energy gained:
5500 − 800 = 4700 J
Therefore:
ΔU = +4700 J
This "energy in minus energy out" method can be particularly useful when several transfers occur.
Connecting Energy Balances to P–V Diagrams
A P–V diagram can provide information about the work term in the First Law.
Recall:
area under a P–V curve = work done by the gas
Suppose the area under an expansion curve represents:
W = 1200 J
and 2,000 J enters by heating.
Then:
ΔU = Q − W
ΔU = 2000 − 1200
ΔU = +800 J
This connects graphical analysis directly to conservation of energy.
Energy Conservation in a Complete Cycle
If a thermodynamic system completes a cycle, it returns to its original state.
Therefore:
U_final = U_initial
and:
ΔU = 0
From:
ΔU = Q − W
we obtain:
Q_net = W_net
So over a complete cycle, the net energy transferred to the system by heating equals the net work done by the system.
This principle is central to the operation of heat engines.
Checking an Energy Balance
After completing a calculation, check that:
total energy entering = total energy leaving + increase in stored energy
For example:
Energy entering = 6,000 J
Work leaving = 2,200 J
Increase in internal energy = 3,800 J
Check:
6000 = 2200 + 3800
6000 = 6000
The energy balance is satisfied.
This is one of the best ways to catch mistakes.
Common Mistakes
Not defining the system
Whether energy is entering or leaving depends on what has been chosen as the system.
Confusing work done by and work done on the gas
Using our convention:
- work done by gas → W positive
- work done on gas → W negative
Ignoring energy losses
Real engineering systems transfer energy through heating, friction, sound and other pathways.
Assuming inefficiency violates conservation of energy
Energy is never "lost" in the conservation sense. It is transferred into less useful forms or to the surroundings.
Mixing sign conventions
If using:
ΔU = Q − W
then W means work done by the system. Keep this convention throughout the calculation.
Ignoring assumptions
A calculated result may only be valid if assumptions such as constant pressure, negligible heat loss or ideal-gas behaviour are reasonable.
Key Terms
Energy balance — an accounting of all energy entering, leaving, and being stored within a system.
System — the object or collection of matter selected for analysis.
Surroundings — everything outside the system.
System boundary — the boundary separating the system from its surroundings.
Heat engine — a device that converts part of an energy transfer from a high-temperature source into mechanical work.
Compressor — a device that uses mechanical work to increase the pressure of a gas.
Turbine — a machine that extracts energy from a moving or expanding fluid to produce mechanical work.
Efficiency — the ratio of useful energy output to total energy input.
Assumption — a simplifying condition adopted when constructing or using a model.
Adiabatic process — a process in which no energy is transferred by heating.
Ideal gas — a simplified model of a gas in which molecular volume and intermolecular interactions are neglected.
Energy loss — an informal description of energy transferred away from the desired output; the energy itself is still conserved.
Key Takeaways
- The First Law provides an energy balance for thermodynamic systems.
- Energy entering, leaving, and being stored must all be accounted for.
- Using our convention, ΔU = Q − W.
- Engines transfer part of their energy input into useful mechanical work.
- Compressors require mechanical work to be transferred into a gas.
- Turbines extract energy from expanding or moving fluids.
- Refrigerators use work to transfer energy from a cooler region to warmer surroundings.
- Efficiency measures the fraction of input energy converted into useful output.
- An inefficient machine still obeys conservation of energy.
- Engineering calculations often rely on simplifying assumptions.
- Important assumptions can include negligible heat transfer, constant pressure, ideal-gas behaviour, and negligible friction.
- P–V diagrams can provide the work term needed for an energy balance.
- For a complete thermodynamic cycle, ΔU = 0.
- Every realistic energy analysis should ask: Where does the energy enter, where does it leave, and where is it stored?
Check Your Understanding
- A gas receives 2,800 J by heating and does 900 J of work. Calculate ΔU.
- A gas does 1,200 J of work while its internal energy decreases by 300 J. Determine Q.
- A compressor does 2,500 J of work on a gas while 600 J leaves by heating. Calculate the change in internal energy.
- A heat engine receives 10,000 J per cycle and transfers 6,500 J to its surroundings. Calculate its useful work output.
- Calculate the efficiency of the engine in Question 4.
- An engine receives 15,000 J and has an efficiency of 32%. Calculate its useful work output.
- A refrigerator removes 600 J from its interior and receives 250 J of electrical work. How much energy is transferred to the room?
- Explain why a compressor can become hot while operating.
- A gas receives 3,000 J by heating. The area under its P–V expansion curve represents 1,100 J of work. Calculate ΔU.
- Explain why an engine with an efficiency of 30% does not violate conservation of energy.
- A calculation assumes that a gas expansion is adiabatic. Explain what this assumption means and give one situation in which it might be approximately reasonable.
- An engineer models a compressor as frictionless and perfectly insulated. Identify two ways in which a real compressor would differ from this idealized model.