Unit 01 · Lesson 07 of 08

Energy crosses. Heat and work describe how.

A warmer wall, a rotating shaft, and a moving piston can all transfer energy. The mechanism at the boundary determines whether we call the transfer heat or work.

Identify a transfer mechanism

What you’ll be able to do

Identify heat, electrical work, shaft work, and boundary work.
Apply a consistent work sign convention.
Recognize process dependence through a pressure-volume diagram.

Before you touch the controls

Can two processes with the same initial and final volumes do different amounts of boundary work?

Explore the modelChange a control. Follow the response.

Work done by the gas200 J

Same starting state (1 L, 300 kPa) and ending state (3 L, 100 kPa). Different shaded area.

Reduce pressure at fixed volume, then expand at 100 kPa. The constant-volume segment does no boundary work.

Quasi-equilibrium piston paths; only pressure-volume work shown. 1 kPa· $\mathrm{L} = 1 \mathrm{J}$ .

Heat is transfer driven by a temperature difference.

Energy transferred across a boundary because of a temperature difference is heat. The system stores internal energy; it does not store heat as a property. Conduction, convection, and radiation are heat-transfer mechanisms you will examine later.

Work has several mechanisms.

A shaft turning against a torque transfers work. Electrical current crossing a potential difference transfers work. A piston moving against pressure transfers work. Throughout these lessons Q is positive into a closed system and W is positive when done by it.

\Delta E=Q-W\quad\text{(closed system)}

Moving-boundary work depends on the path.

For a quasi-equilibrium piston process, boundary work is the integral of pressure over volume change. Expansion gives positive work under the work-by-positive convention; compression gives negative work. At constant pressure this reduces to pΔV. The area under a p–V path measures the work.

W_b=\int_{V_1}^{V_2}p\,dV,\qquad W_b=p(V_2-V_1)\quad\text{(constant pressure)}

Know when the diagram applies.

The quasi-equilibrium model assumes sufficiently well-defined intermediate states and negligible pressure imbalance at the moving boundary. For rapid, irreversible motion, do not automatically integrate a guessed uniform gas pressure. The work must be evaluated using the appropriate boundary forces and displacement.

A worked example

A slow expansion at 200 kPa.

A gas expands from 1 L to 3 L at constant pressure. $\Delta V = 2$ $\mathrm{L} = 0.002$ m³.
Work by the gas is $200,000 \times 0.002 = 400 \mathrm{J}$ . Equivalently, 1 kPa· $\mathrm{L} = 1 \mathrm{J}$ .
If 700 J of heat enters and changes in KE and PE are negligible, $\Delta U = 700 - 400 = 300 \mathrm{J}$ .

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

Read the area under the path.

A gas expands quasi-statically from 1 L to 3 L at 200 kPa. Find boundary work. Use the visual to compare other final volumes.

A small hint

A rectangle’s area is $pressure \times volume$ change.

Explore the visual solution

Boundary work by the gas400 J

200 kPa × ( $3 - 1$ ) L. Expansion work is positive.

What if you change it?Final volume3 L

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

The process has constant pressure, so $W=p(V_2-V_1)$ .

Carry this forward. The p–V area is work only with the appropriate process and boundary-work assumptions.

02 · Reason

A rigid tank still receives work.

An insulated, rigid, sealed tank contains a resistor operating at 12 V and 2 A for 30 s. Find the total energy increase of tank contents plus resistor. Neglect all other transfers.

A small hint

Electrical power is $voltage \times current$ ; zero boundary work is not zero total work.

Explore the visual solution

Electrical energy input720 J

12 $V \times 2$ $A \times 30$ s. Internal-energy gain equals the 24 W input multiplied by time.

What if you change it?Operating time30 s

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Electrical $power = 12 \times 2 = 24 \mathrm{W}$ ; energy $input = 24 \times 30 = 720 \mathrm{J}$ .

Carry this forward. Name each possible work mechanism before neglecting work.

03 · Challenge

Same endpoints, different work.

Two quasi-equilibrium paths go from (1 L, 300 kPa) to (3 L, 100 kPa). Path A lowers pressure at 1 L, then expands at 100 kPa. Path B expands at 300 kPa, then lowers pressure at 3 L. Compare their work.

A small hint

A constant-volume segment contributes no boundary work.

Explore the visual solution

Boundary work by the gas200 J

The shaded area differs even though both end states match. Vertical path segments contribute zero work.

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

Path A: $\mathrm{W} = 100$ × ( $3 - 1$ ) = 200 J. The vertical segment contributes zero.

Carry this forward. Work and heat depend on the process; internal-energy change depends only on the endpoints.

Your learning path

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