First Law Budget - Heat & Thermodynamics Step by Step

The first law tracks where transferred energy goes in a process. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Start with a heat input

A process receives 20 joules of heat. Some energy leaves as work and the rest stays as internal energy.

Q_{\text{in}} = 20\ \text{J}

Split the energy

The process does 12 joules of work and stores 8 joules as internal energy.

Q_{\text{in}} = W + \Delta U

Check the budget

Adding 12 joules and 8 joules gives the original 20 joules.

Q_{\text{in}} = 12\ \text{J} + 8\ \text{J} = 20\ \text{J}

More work means more input if internal change is fixed

Hold the internal-energy change at 8 joules. The diagram shows the last row; more work requires more input energy.

\begin{array}{c|c|c}W&\Delta U&Q_{\text{in}}\\4\ \text{J}&8\ \text{J}&12\ \text{J}\\8\ \text{J}&8\ \text{J}&16\ \text{J}\\12\ \text{J}&8\ \text{J}&20\ \text{J}\\\end{array}

More internal storage also raises input

Now hold work at 12 joules. More stored internal energy also requires more heat input.

\begin{array}{c|c|c}W&\Delta U&Q_{\text{in}}\\12\ \text{J}&4\ \text{J}&16\ \text{J}\\12\ \text{J}&8\ \text{J}&20\ \text{J}\\12\ \text{J}&12\ \text{J}&24\ \text{J}\\\end{array}

The same input can split different ways

For a fixed 20 joule input, the work and internal-energy rows must still add to the same total. The diagram shows the middle row.

\begin{array}{c|c|c}W&\Delta U&Q_{\text{in}}\\4\ \text{J}&16\ \text{J}&20\ \text{J}\\12\ \text{J}&8\ \text{J}&20\ \text{J}\\16\ \text{J}&4\ \text{J}&20\ \text{J}\\\end{array}