Work and heat in simulations

Work and heat are two ways energy crosses a system boundary. Work is transfer associated with organized generalized forces and displacements. Heat is transfer caused by a temperature difference. Neither is a substance stored inside the system; once transferred, the energy contributes to the system's state, including internal energy.

Choose one first-law convention

This lesson takes heat entering the system as positive and work done by the system on its surroundings as positive. The first law is

\[\Delta U=Q-W\]

Here \(U\) is a state function. \(Q\) and \(W\) describe a process. If a system absorbs \(10\,\mathrm J\) and does \(3\,\mathrm J\), then

\[\Delta U=10\,\mathrm J-3\,\mathrm J=7\,\mathrm J\]

Many engineering texts define work done on the system as positive. They write

\[\Delta U=Q+W_{\mathrm{on}},\qquad W_{\mathrm{on}}=-W\]

The physics agrees. Errors appear when a formula from one convention is combined with a sign label from the other.

Pressure–volume work

For a piston with cross-sectional area \(A\) moving outward by \(dx\) against external pressure \(p_{\mathrm{ext}}\), the volume change is \(dV=A\,dx\) and the work done by the gas is

\[\delta W=p_{\mathrm{ext}}\,dV\]

For a finite process,

\[W=\int_{V_i}^{V_f}p_{\mathrm{ext}}\,dV\]

During a quasistatic mechanical process, the gas pressure differs only infinitesimally from external pressure, so a measured equilibrium path \(p(V)\) can be used in the integral. During a rapid irreversible expansion, the system may not have one uniform pressure. Replacing external pressure by an imagined equilibrium gas curve can give the wrong work.

Expansion has \(dV>0\) and positive work by the system. Compression has \(dV<0\) and negative work by the system, so it tends to raise internal energy when heat is absent.

Work depends on the path

On a pressure–volume diagram, quasistatic work is the signed area under the path. Initial and final points alone do not determine it. At constant pressure \(p_0\),

\[W=p_0(V_f-V_i)\]

At constant volume, \(W=0\) even if pressure and temperature change. For an isothermal ideal-gas process, \(p=nRT/V\) and

\[W=nRT\ln\left(\frac{V_f}{V_i}\right)\]

This logarithmic result requires fixed temperature, fixed amount of ideal gas, and a quasistatic path. It is not a generic endpoint work formula.

Numerically integrating a measured path

For sampled quasistatic points ordered along the process, the trapezoidal approximation is

\[W\approx\sum_{j=0}^{M-2}\frac{p_j+p_{j+1}}{2}(V_{j+1}-V_j)\]

import numpy as np

volumes = np.linspace(0.01, 0.02, 101)
external_pressures = np.full_like(volumes, 100_000.0)
work_by = np.trapezoid(external_pressures, volumes)
print(f"Expansion work by the gas: {work_by:.2f} J")

Pressure times volume has units \(\mathrm{Pa\,m^3}=\mathrm J\). Keep path order: sorting points by volume can destroy loops and reverse parts of the actual process. Repeat the integration on finer measurements or against an analytic path to estimate quadrature error.

What counts as heat

Heat transfer occurs because of a temperature difference and can proceed by conduction, convection, or radiation. The symbol \(Q\) denotes the net energy transferred by these mechanisms over the chosen interval. A body at equilibrium has internal energy, not a stored quantity of “heat.”

For a material with approximately constant specific heat \(c\) and no phase change,

\[Q=mc(T_f-T_i)\]

is useful under the stated conditions. It is not the definition of heat and may fail when \(c\) varies strongly, work occurs, or a phase transition absorbs latent energy. During melting or boiling, energy can enter while temperature remains nearly constant.

Heat transfer rate may be modeled with a law such as Newton cooling,

\[\dot Q=hA(T_{\mathrm{environment}}-T_{\mathrm{system}})\]

but this is an empirical lumped model. The coefficient \(h\) depends on geometry, flow, and material conditions; it is not a universal constant.

Adiabatic does not mean isothermal

An adiabatic boundary permits no heat transfer, so \(Q=0\). The first law becomes

\[\Delta U=-W\]

An expanding ideal gas does positive work and usually cools because internal energy decreases. A compressed gas can warm. Adiabatic therefore does not mean constant temperature. Isothermal means constant temperature and can require heat exchange to balance work.

For a quasistatic adiabatic ideal gas with constant heat capacities,

\[pV^\gamma=\mathrm{constant}\]

where \(\gamma=C_P/C_V\). This path relation does not describe every adiabatic irreversible process.

Microscopic simulations and thermostats

In molecular dynamics, moving walls do work through particle collisions. A thermostat algorithm exchanges energy with a modeled environment. Rescaling velocities is not energy-neutral: multiplying velocities by a factor \(s\) multiplies kinetic energy by \(s^2\).

velocities = np.array([-2.0, -1.0, 1.0, 2.0])
energy_before = 0.5 * np.sum(velocities**2)
rescaled = velocities * 1.1
energy_after = 0.5 * np.sum(rescaled**2)
thermostat_transfer = energy_after - energy_before
print("Energy added by rescaling (scaled units):", thermostat_transfer)

Whether thermostat transfer is called heat depends on the model and thermodynamic interpretation. At minimum, record it as a named external transfer. A numerical damping term that removes energy without a ledger is not a physical thermal model.

Friction also needs a boundary decision. For a block-only system, energy transferred to a surface crosses the boundary. For a block-plus-surface system, friction can redistribute energy internally. Changing the boundary changes classification, not total conservation.

The ledger and its residual

Rates need the same sign discipline. Power done by the system is \(\dot W\), while thermal power entering is \(\dot Q\). For a closed system with no resolved changes in bulk kinetic or external potential energy, the rate form is

\[\frac{dU}{dt}=\dot Q-\dot W\]

In a discrete update, integrate rates over the same time interval used to change stored energy. Multiplying a rapidly changing endpoint rate by the full step can create a large balance error. A midpoint or trapezoidal rule is usually more consistent for smooth data, while events that occur inside a step may require splitting the interval.

Also distinguish energy from power in output. Joules describe an accumulated transfer; watts describe joules per second. A graph labeled “heat” but storing instantaneous watts cannot be subtracted directly from an internal-energy change. Integrate it first, preserve timestamps, and state the quadrature rule.

For each update, record

\[R=\Delta U-Q+W\]

The residual \(R\) should be near zero within measurement and numerical error after every modeled transfer is included.

initial_internal, final_internal = 2000.0, 2500.0
heat_in = 1500.0
residual = final_internal - initial_internal - heat_in + work_by
assert np.isclose(residual, 0.0)
print("First-law residual (J):", residual)

Do not force the residual to zero by assigning it to heat. Investigate omitted boundary flux, inconsistent signs, changing potential-energy zeros, integration error, or missing energy stores.

Verification cases

With a rigid boundary and \(10\,\mathrm J\) of heat input, require \(\Delta U=10\,\mathrm J\). For an adiabatic expansion that does \(3\,\mathrm J\), require \(\Delta U=-3\,\mathrm J\). For adiabatic compression with \(W=-3\,\mathrm J\), require \(\Delta U=3\,\mathrm J\).

Reverse a stored process path and confirm that signed work reverses. Compare sampled pressure–volume integration with constant-pressure and ideal isothermal analytic results. Refine time steps and path sampling, and require the first-law residual to converge.

Rapid nonequilibrium processes can have gradients, shocks, viscosity, chemical reactions, and moving interfaces. They may require local energy equations rather than one uniform temperature and internal-energy scalar. The ledger remains necessary, but every term must match the richer model.

A worked Python notebook

One mole of ideal gas expands or compresses reversibly at 300 kelvin from 0.01 cubic metres. The question for this calculation is precise: how does isothermal work by gas change when we change final volume, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

This formula requires the external pressure to follow the quasistatic gas pressure. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 0.005 to 0.03 m^3. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for final volume, expressed in m^3. The output is isothermal work by gas, expressed in J. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = 2494.3387854 \log{\left(100.0 q \right)}\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 0.02 m^3, the expression gives 1728.94 J. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("8.314462618*300*log(q/0.01)", locals={"q": q})
reference_input = 0.02
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "m^3")
print("Predicted isothermal work by gas:", reference_output, "J")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives -1728.94 J. The midpoint, at 0.0175 m^3, gives 1395.87 J. The upper endpoint gives 2740.31 J. Read these together with the assumptions: they describe different choices of final volume within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([0.005, 0.017499999999999998, 0.03])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} m^3 -> {output_value:.6g} J")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows final volume; the vertical axis shows isothermal work by gas. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(0.005, 0.03, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Final volume (m^3)")
ax.set_ylabel("Isothermal work by gas (J)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does isothermal work by gas change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 0.02 m^3 is 124717 in those ratio units. Locally, the output increases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = \frac{2494.3387854}{q}\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 0.020125
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "J")
print("Full expression:", full_prediction, "J")
print("Approximation difference:", full_prediction - linear_prediction, "J")

Here the changed input is 0.020125 m^3. The tangent estimate is 1744.53 J, while the full expression gives 1744.48 J. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 0.024999999999999998
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of final volume inside the stated interval, predict whether isothermal work by gas will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: This formula requires the external pressure to follow the quasistatic gas pressure. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of J: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(0.03) - evaluate(0.005))
for points in [101, 201, 401]:
    grid = np.linspace(0.005, 0.03, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

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