Use conservation to calculate what changes, then determine a shared temperature from heat capacities.
By the end: apply a consistent work sign and solve a finite-object thermal contact problem.
Write an energy account
The first law is energy conservation applied to a chosen system. Here Q is net heat entering the system, W is net work done by the system, and ΔU is the final internal energy minus its initial value. With these signs, subtract work output from heat input. If a gas receives 100 J of heat and pushes out a piston with 40 J of work, its internal energy rises by 60 J.
Compression reverses the work sign. If the surroundings do 40 J of work on an insulated gas, W is −40 J and Q is zero, so internal energy rises by 40 J. Heating and compression can reach the same final internal energy by different routes. U is a state function: its change depends on endpoints. Q and W describe transfers along a route, not properties that a state stores.
Assign signs before adding transfers
Choose the gas as the system, then write every transfer with the course's heat-in and work-out convention. A heat arrow entering has positive Q. A mechanical arrow leaving has positive W. The first law subtracts W because that energy leaves the gas. Suppose one stage admits 900 J as heat and delivers 300 J as work. Subtract 300 from 900 to get a 600 J internal-energy increase. The two transfers need not be equal, and neither arrow represents a quantity stored inside the final gas. After the stage, the gas has a changed U, not a separate stored Q and W.
Next let the same gas release 200 J as heat while receiving 100 J of compression work. Heat leaving means Q is negative 200 J. Work entering means W is negative 100 J. Subtraction of that negative work adds 100 J, giving an internal-energy change of negative 100 J for this stage. Across both stages, net heat is 700 J and net work output is 200 J. Their difference is 500 J, matching the sum of the separate changes, 600 J minus 100 J.
The stage table keeps these additions exact. It also exposes a common error in the phrase total heat input. Gross inward heat here is 900 J, but net Q is only 700 J after the outward transfer. Similarly, gross outward work is 300 J, while net W is 200 J after compression. Use net signed transfers in the first law. If a question instead asks for gross energy entering or leaving by one route, name that quantity explicitly rather than replacing the signed net without notice.
Add signed stages, then check their sum
Stage
Q, heat in
W, work out
ΔU
Heat and expansion
+900 J
+300 J
+600 J
Cooling and compression
−200 J
−100 J
−100 J
Net over both stages
+700 J
+200 J
+500 J
Q counts heat inward; W counts work outward. Every row obeys ΔU = Q − W. Entries are joules for the chosen gas, with no bulk kinetic-energy or gravitational-energy change.
Mechanical work has a geometric meaning. For a piston moving 0.05 m against a constant opposing force of 2,000 N, the work output is force times displacement, or 100 J. If the piston area is 0.01 m², the external pressure is 200,000 Pa and the volume increase is 0.0005 m³. Multiplying pressure by volume change gives the same 100 J. The pressure is absolute opposing pressure, and the work account includes all of that opposition rather than only the part above ambient pressure. A gauge reading would omit the atmospheric contribution unless that contribution were explicitly accounted for elsewhere.
For this geometry, follow the displacement rectangle and then the pressure-volume rectangle. Both areas encode the same work using different quantities. No claim that gas pressure equals external pressure is needed for constant external-pressure boundary work. Using an equilibrium gas-pressure curve instead requires a quasistatic mechanical description. Reversibility needs more, including the absence of friction and finite driving differences. Slow piston motion alone does not prove it. The energy balance itself remains valid even when rapid motion prevents assigning a uniform gas pressure throughout the intermediate states.
Piston displacement gives the same work as a pressure-volume area
Left: piston displacement is schematic, not a length scale. Right: the shaded rectangle uses absolute external pressure, 200 kPa, and a volume increase of 0.5 L. Its area is 100 J because 1 kPa L = 1 J. No gas-pressure path or reversible-process assumption is drawn.
Connect energy to a temperature change
Heat capacity C is the energy required per kelvin of temperature rise under a specified constraint. Specific heat capacity c is per kilogram, and total capacity is mass times specific capacity. For a body doing negligible work, with constant capacity and no phase change, heat input equals C times the temperature change. A 100 J/K body needs 1,000 J to warm by 10 K. Gases require extra care because expansion can use part of the heat.
Put a 400 K block of capacity 100 J/K in contact with a 300 K block. Insulate the pair and allow no work or phase change. At the common final temperature, the hot block’s lost energy must equal the cold block’s gain. Rearranging that equality gives the capacity-weighted average below. Equal capacities give 350 K; a cold block with four times the capacity gives 320 K, closer to its initial 300 K.
Derive the common temperature from two changes
Let x be how many kelvin the initially cold block warms in the contact model. Its final temperature is 300 K plus x. Since the initially hot block starts at 400 K and both end together, the hot block cools by 100 K minus x. With hot capacity 100 J/K and cold capacity 200 J/K, conservation requires 100 times the quantity 100 minus x to equal 200 times x. This equation says equal transferred energies, not equal temperature changes, and keeps the initial temperature gap visible. Both sides must have joule units before their numerical values can be equated in this way.
Expand the left side to obtain 10,000 minus 100x equals 200x. Move the negative 100x to the right, giving 10,000 equals 300x. Thus x is about 33.33 K and the final temperature is about 333.33 K. The hot block cools by about 66.67 K. Using unrounded values, either capacity multiplied by its temperature change gives 6,666.67 J transferred. Round only after comparing the two accounts. Multiplying separately rounded changes can otherwise create a small apparent mismatch that is only arithmetic rounding.
The diagram represents each temperature change as a horizontal interval and its capacity as a rectangle height. Its area is energy in joules because J/K multiplied by K gives J. The colder block has twice the rectangle height but half the interval width. Equal areas show energy conservation without requiring equal widths. Halving the colder block's rise compensates exactly for doubling its energy requirement per kelvin of rise. This also explains the weighted average. A large capacity permits only a small temperature change for a given transfer, so the common temperature sits nearer the starting temperature of the larger-capacity body.
Before changing the contact controls, predict the direction of the result. Select the equal-capacity preset and check that each block changes by 50 K. Then select the cold block with four times the capacity. Predict whether the final temperature moves above or below the equal-capacity marker, and whether the transferred energy increases or decreases. The displayed 320 K endpoint means the hot block cools by 80 K and transfers 8,000 J. The cold block warms by only 20 K, but its 400 J/K capacity absorbs that same amount.
Equal energy areas need not have equal temperature widths
Both rectangles use 3 horizontal SVG units per kelvin and 0.65 vertical units per J/K. Each area represents 6,666.67 J, rounded. Width is the magnitude of temperature change, not absolute temperature; height is whole-sample capacity.
Use the slider to select 300 J/K for the cold block. Do not estimate the answer by averaging the two neighboring preset temperatures. Solve 100 times the quantity 100 minus x equals 300x instead. This gives x equal to 25 K, a final temperature of 325 K, and 7,500 J transferred. Check all three against the model output. The dependence is curved because capacity appears in the denominator of the weighted average. Doubling cold capacity therefore does not halve the entire final temperature or double the heat passed.
Capacity changes the weighted average along a curve
Only cold capacity changes. Hot capacity stays 100 J/K; starting temperatures stay 400 K and 300 K. The vertical scale is cropped to 300–350 K. Labeled points are algebraic endpoints, not measured thermal histories.
Work through the mechanism
First law for the chosen closed system
ΔU, Q, and W are in joules. Work input makes W negative. Overall motion and changes in height are excluded here.
Common temperature after thermal contact
C values are constant heat capacities in J/K, not temperatures. T values are in kelvin. No work, losses, or phase change occur.
Let two finite blocks reach a common temperature
Hot block: 100 J/K at 400 K. Cold block starts at 300 K.
Final temperature 333.33 K. The hot block loses 6,666.67 J; the cold block gains the same amount.
Case
Cold capacity
Final T
Heat passed
Reference
100 J/K
350 K
5,000 J
Selected
200 J/K
333.33 K
6,666.67 J
The ring marks the fixed equal-capacity reference; the point marks the selected final temperature. The scale spans 300–400 K, not zero. Capacities are constant; no work, losses, or phase changes occur. Controls replace only the cold capacity, with no time model or playback.
Cold capacity
Final temperature
Heat transferred
100 J/K
350 K
5,000 J
200 J/K
333.33 K
6,666.67 J
400 J/K
320 K
8,000 J
Hot capacity is always 100 J/K; initial temperatures are 400 K and 300 K. Values are rounded.
Keep the endpoint balance separate from its time history
The contact model chooses an equilibrium endpoint, not a sequence of real temperatures over measured seconds. A narrow conducting connection and a broad conducting connection can lead to the same final balance when the pair is isolated and their capacities stay unchanged. They need not reach it at the same rate. The temperature difference establishes transfer direction without specifying joules transferred per second. The calculation contains capacities and starting temperatures, but no conductivity, contact area, or time parameter. Changing cold capacity replaces a hypothetical sample property in the balance. It does not specify a physical adjustment made during a timed contact process.
To reason about intermediate states without inventing a clock, let q be the positive energy already transferred from the hot block. In the equal-capacity case, the hot temperature is 400 K minus q divided by 100 J/K. The cold temperature is 300 K plus q divided by 100 J/K. At q equal to 2,000 J these are 380 K and 320 K. At 5,000 J both reach 350 K. This progress coordinate counts transferred joules; equal spacings along it do not represent equal elapsed times.
For cold capacity 400 J/K, use the same hot expression but divide q by 400 J/K for the cold rise. At 5,000 J, the hot block is at 350 K and the cold block at 312.5 K. They have not equilibrated. At 8,000 J the temperatures meet at 320 K. Compare that intersection with the equal-capacity intersection in the diagram. A given hot-block loss produces different cold-block rises because the capacity changed. Continuing these straight lines beyond their intersection would describe an imposed transfer, not spontaneous contact equilibration.
Conservation alone does not tell us which way an isolated contact can proceed. Start again with two 100 J/K blocks at 400 K and 300 K. Transferring 5,000 J from the hotter block gives 350 K for both, with energy changes of minus 5,000 J and plus 5,000 J. Now write a different paper proposal: move 1,000 J from the colder block to the hotter one. The proposed temperatures become 290 K and 410 K. Their energy changes also sum to zero. Both accounts conserve energy, although only the first describes ordinary passive equilibration.
Transferred energy locates the meeting point without supplying a clock
Both panels use the same q and temperature scales. Solid lines show the hot block; dashed lines show the cold block. Lines stop at equality. q is accumulated energy passed, not time. Hot capacity is always 100 J/K.
The second proposal would make the temperature difference grow without another change. The first-law arithmetic does not reject it because that law tests conservation, not the direction of an allowed process. A separate thermodynamic constraint is needed. The later entropy lesson develops that constraint after introducing its quantities and calculation. Here, keep the questions distinct. The capacity balance finds the common temperature if equilibrium is reached under the stated conditions. A transfer model explains the approach and its timing. Neither a balanced energy sum nor a chosen progress coordinate by itself supplies all that information.
Conservation alone does not determine transfer direction
Proposed transfer
Hot-block ΔU / final T
Cold-block ΔU / final T
Combined ΔU
5,000 J hot to cold
−5,000 J / 350 K
+5,000 J / 350 K
0 J
1,000 J cold to hot
+1,000 J / 410 K
−1,000 J / 290 K
0 J
Both blocks have constant capacity 100 J/K and start at 400 K and 300 K. Each proposed account conserves energy. The second is not passive equilibration; it would need another interaction to be physically achievable.
Do not confuse an endpoint with a rate
Heat capacity determines how far a temperature changes for a given energy transfer. It does not determine how fast the transfer happens. Contact area, thermal conductivity, and surface heat exchange affect the time needed to approach equilibrium. The model calculates the final balance and makes no cooling-time prediction.
An isolated total can have constant internal energy while its parts change temperature. The final average is valid only under the stated assumptions. Heat leaking into the surroundings, a moving boundary, temperature-dependent capacities, a phase transition, or a chemical reaction requires a different energy account. A body’s temperature can even stay constant while its energy changes during melting.
Test what the capacity shortcut leaves out
Whole-sample capacity is not automatically a universal multiplier relating net heat to temperature change. For a solid block doing negligible work in one phase, constant capacity makes that shortcut useful. For an expanding gas, some incoming heat leaves as work. Use a stipulated sample whose internal energy rises by 150 J for a 30 K temperature rise in each of the following cases. At fixed volume, with no other work, all 150 J must arrive as heat. Its heat input per kelvin is therefore 5 J/K under this constraint. This energy-versus-temperature assumption is part of the example, not a rule for every material.
For the second case, let the sample push a frictionless piston against a constant opposing pressure of 100 kPa while its volume increases by 1 L. Pressure times volume change gives 100 J of work, since one kilopascal times one litre equals one joule. Adding this work to the stipulated 150 J internal-energy rise requires 250 J of heat input. For the same 30 K rise, that is about 8.333 J/K. These are whole-sample capacities, not per-mole values. Using the fixed-volume value for this heating process would omit the energy delivered to the surroundings. The ideal-gas lesson later explains a material model consistent with these stipulated changes.
The energy strips compare these routes on one joule scale. Both end with the same internal-energy increase, but they do not end at the same volume and pressure because their constraints differ. To isolate route dependence between identical endpoint states, consider instead two permitted accounts with the same final U. One route supplies 500 J as heat with zero work. Another supplies 800 J as heat and outputs 300 J as work. Both increase U by 500 J. The balance permits these accounts; a material and process description must establish the endpoints.
Fixed-pressure heating supplies internal energy and work
Horizontal lengths share a zero-based joule scale. Stipulated internal-energy rise 150 J and temperature rise 30 K in both cases. The expanding case does 100 kPa × 1 L = 100 J of work. Constraints differ, so final volumes and pressures are not identical.
A further contrast is phase change. During melting at a specified coexistence pressure, added energy can increase the liquid mass fraction while temperature remains fixed. Then dividing transferred heat by the measured temperature rise would not yield a finite single-phase capacity. Nor does the liquid fraction mean the same fraction of the sample's volume, because solid and liquid densities differ. The missing variable is how much mass changed phase. A contact calculation spanning such a transition must include the phase-energy change before solving for the common final state.
Finally, examine a hypothetical container omitted from a two-block account. Let the hot block have capacity 100 J/K at 400 K, the cold block 100 J/K at 300 K, and the container 200 J/K at 300 K. If all three equilibrate in an isolated assembly, the total initial capacity-weighted temperature is 130,000 J and the total capacity is 400 J/K. The endpoint is 325 K, not 350 K. Check that the hot block loses 7,500 J while the cold block gains 2,500 J and the container gains 5,000 J.
A container can receive energy inside the isolated boundary
Schematic enclosure and internal transfers. The container participates with constant capacity 200 J/K and starts at 300 K. No energy crosses the outer insulated boundary. All final temperatures are 325 K; arrows encode direction, not scaled rate.
Exercises
Handle work input
An insulated closed system receives 250 J of mechanical work. Its overall motion and height do not change. What are Q, W, and ΔU in this course’s convention?
Hint
W counts work done by the system, not work done on it.
Worked answer
Quantity
Value
Heat entering Q
0 J
Work done by system W
−250 J
Internal-energy change Q − W
+250 J
Work input carries a negative work-output sign.
Q is zero, W is −250 J, and ΔU is +250 J. Insulated does not mean that energy cannot enter as work. Whether the temperature rises, and by how much, requires a material and state model.
Calculate a new contact temperature
A block of capacity 200 J/K starts at 360 K. A second block of capacity 100 J/K starts at 300 K. Find the equilibrium temperature and heat transfer for their isolated, rigid pair with constant capacities.
Hint
Weight each starting temperature by capacity, then check both energy changes.
Worked answer
Step
Calculation
Weighted temperature
(200 × 360 + 100 × 300) / 300 = 340 K
Hot block loses
200 × 20 = 4,000 J
Cold block gains
100 × 40 = 4,000 J
Equal transferred energies do not imply equal temperature changes.
The weighted temperature is 340 K. The first block cools by 20 K and loses 4,000 J. The second warms by 40 K and gains 4,000 J. The hotter block’s larger capacity keeps the final temperature closer to 360 K than to 300 K.
Calculate the common temperature
Capacities use J/K and temperatures use K. The common capacity unit cancels. Then check both energy changes separately.
Sources and further reading
These optional references support the stated laws and model assumptions. The numerical examples and diagrams are authored teaching models, not experimental measurements or engineering specifications. Selected relevant passages were reviewed, not entire textbooks or a comprehensive research literature.