Choose the boundary, name the transfers, and identify the energy stores. The first law makes the story quantitative; efficiency asks how much of the input serves your intended purpose.
Bring the ideas together
What you’ll be able to do
Apply ΔE=Q−W with a stated sign convention.
Include kinetic and potential energy changes when needed.
Distinguish conservation from useful-output efficiency.
Before you touch the controls
If a system loses heat while being compressed, must its internal energy fall?
Explore the modelChange a control. Follow the response.
Q (heat into system)-30 J
W (work by system)-80 J
ΔU (internal-energy change)50 J
ΔU=Q−W−ΔKE50 J
Closed system; ΔPE=0. A negative W is work input. The piston and arrows indicate transfer direction; geometry is schematic.
01
Write the balance before inserting numbers.
For a closed system, the first law is ΔU+ΔKE+ΔPE=Q−W. Here Q is heat into the system and W is work done by it. Work input is therefore negative W. A control volume needs additional energy terms associated with flowing mass; do not silently apply this fixed-mass equation to a pump.
ΔU=Q−W−ΔKE−ΔPE
02
Signs carry physical meaning.
Compression can increase internal energy even while heat leaves. The net outcome depends on the magnitudes. If Q=−30J and W=−80J, the total stored energy rises by 50 J. A negative transfer is not an error; it indicates a direction opposite your chosen positive convention.
03
Useful is a design decision.
For an ordinary work-output device such as a motor or pump, energy efficiency is useful output divided by input. The remainder is not destroyed; it appears in other energy transfers or stores. A heat pump uses a different metric, coefficient of performance, because its purpose is to move heat rather than convert input work entirely into heat.
η=EinEuseful,out
04
A cycle returns to its initial state.
When a system completes a cycle, all its state properties return to their initial values, so the net change in stored energy is zero. For a closed-system cycle, net heat transfer equals net work. The first law enforces accounting but does not by itself guarantee that a proposed process is physically achievable.
A worked example
Cooling and compression at the same time.
A closed gas loses 30 J of heat, so Q=−30J. External compression does 80 J of work on it, so W=−80J.
ΔE=Q−W = −30 − (−80) = +50 J.
If KE and PE are unchanged, ΔU = +50 J. Work input more than compensates for heat loss.
Make the idea your own
Predict. Calculate. Question the model.
Try each problem before opening its solution. Then change a value to test whether the reasoning still holds.
01 · Calculate
Watch the signs during compression.
A closed gas loses 30 J of heat and receives 80 J of compression work. Neglect kinetic and potential energy changes. Find ΔU.
A small hint
Work input is negative when W is defined as work done by the system.
Explore the visual solution
Work input80 J
Heat loss30 J
Internal-energy change50 J
ΔU=Q−W50 J
Q = −30 J and W = −80 J. The minus sign before W turns work input into an energy gain.
What if you change it?
The visual responds to your control. The worked steps below use the original question.
Original question · Step 1 of 3
Q=−30J; W=−80J.
Carry this forward.State the convention, then translate each physical direction into a signed quantity.
02 · Calculate
Where does a motor’s input go?
A motor receives 500 J and delivers 350 J of shaft work. It returns to the same stored-energy state. Find its energy efficiency and the energy transferred elsewhere.
A small hint
Use useful shaft output in the numerator.
Explore the visual solution
Useful shaft output350 J
Other energy transfers150 J
Energy efficiency70%
Input stays at 500 J. Useful output+othertransfers=input when stored energy is unchanged.
What if you change it?
The visual responds to your control. The worked steps below use the original question.
Original question · Step 1 of 3
Efficiency=350/500=0.70=70%.
Carry this forward.Efficiency measures the desired outcome; conservation accounts for every outcome.
03 · Challenge
Complete a cycle with a missing transfer.
A closed system completes a cycle. It receives 900 J of heat in one stage, rejects 500 J in another, and has no other heat transfer. Find net work by the system. If one stage requires 100 J of work input, how much work must the other stages deliver in total?
A small hint
For a cycle, ΔE=0. Net work includes both work input and output.
Explore the visual solution
Net heat = net work400 J
Gross output, with 100 J work input500 J
Net work=900−500=400J. Gross output=network+100J.
What if you change it?
The visual responds to your control. The worked steps below use the original question.
Original question · Step 1 of 3
Net heat=900−500=400J. Over the cycle, ΔE=0.
Carry this forward.Net and gross transfers answer different questions. Keep both directions visible.
Your learning path
You’ve reached the end of Unit 1.
Mark this lesson complete when you can explain the main idea and work through its practice. Completion is your choice.
