How do I solve a first law of thermodynamics problem for a closed system?

First law questions are usually easier when you slow down and define the system before using numbers. A closed system has a fixed amount of mass. Energy can cross the boundary as heat or work, but mass does not cross the boundary. Most student errors come from sign convention, units or using the wrong form of the energy balance.

What the question is asking

The question is asking you to apply conservation of energy. For many closed-system assignment problems, changes in kinetic and potential energy are negligible, so the focus is on internal energy. But you should state that assumption rather than silently ignoring terms.

Key concepts to cover

• Closed system: no mass crosses the boundary
• Heat transfer: energy transfer because of temperature difference
• Work: energy transfer by force through distance, such as boundary work
• Internal energy: microscopic energy of the system
• Sign convention: heat added to the system and work done by the system are often treated as positive
• Energy balance: change in energy equals heat in minus work out
• Units: joules, kilojoules, kilograms and kJ/kg must be handled consistently

Suggested answer structure

1. 1Define the system boundary.
2. 2State whether the system is closed.
3. 3Write the first law equation.
4. 4State assumptions about kinetic and potential energy.
5. 5Insert heat and work with the correct sign.
6. 6Solve for the unknown internal energy change.
7. 7Check units and interpret the sign of the answer.

Mini worked example

Suppose a closed piston-cylinder system receives 20 kJ of heat. During expansion, the system does 7 kJ of boundary work on the surroundings. Neglect kinetic and potential energy changes.

Using the convention:

Delta U = Q - W

where Q is heat added to the system and W is work done by the system:

• Q = +20 kJ
• W = +7 kJ
• Delta U = 20 - 7 = 13 kJ

The internal energy increases by 13 kJ. The result is positive because more energy entered as heat than left as work.

If instead 7 kJ of work were done on the system, then W would be negative under this convention:

• Delta U = 20 - (-7) = 27 kJ

That difference is why the wording "by the system" and "on the system" matters.

Common mistakes

• Using work as positive whether it is done by or on the system
• Forgetting to state the sign convention
• Mixing kJ and J in the same calculation
• Treating a closed system as if mass flow terms apply
• Ignoring kinetic and potential energy without saying they are negligible
• Reporting a number without explaining whether internal energy increased or decreased

How to make the answer stronger

Draw a simple boundary around the system and mark arrows for heat and work. Even if the final answer is numerical, the diagram helps prevent sign errors. Also write the general energy balance first, then simplify it. That shows the marker that you understand why terms are removed.

For engineering assignments, unit discipline is part of the method. If the data gives mass and specific internal energy in kJ/kg, multiply by mass to get kJ before comparing with total heat or work.

Related questions

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• How do I apply Bernoulli's equation to a nozzle problem?
• How do I calculate stress, strain and factor of safety?

Academic use note

This guide is for engineering assignment learning support. It is not professional engineering design advice. Use your module's sign convention, property tables and required assumptions.

Sources and further reading

• MIT notes on the first law of thermodynamics
• MIT OpenCourseWare: Work, Heat, and the First Law
• OpenStax University Physics: First Law of Thermodynamics