Physic Labs

Thermal physics

The first law of thermodynamics

The first law applies energy conservation to a thermodynamic system: its internal-energy increase equals heat added plus work done on it.

Heat and work both transfer energy across a system boundary. The first law combines them to determine the change in internal energy, using an explicit sign convention.

ΔU=Q+Won(Q>0:heat into system,  Won>0:work on system)\Delta U = Q + W_{\text{on}}\qquad (Q>0:\text{heat into system},\;W_{\text{on}}>0:\text{work on system})

Definition: Sign convention

Take Q>0Q>0 when heat enters the system and Won>0W_{on}>0 when the surroundings do work on it (for example, compressing a gas). If the gas does work WbyW_{by} on the surroundings, then Won=−WbyW_{on}=-W_{by} and one may write ΔU=Q−Wby\Delta U=Q-W_{by}. These forms are equivalent.

Set heat transfer and compression/expansion work to check the energy balance. The simulation takes WonW_{on} as positive for work on the system.

Special cases

For an adiabatic process, Q=0Q=0, so work changes the internal energy. In an isochoric process the boundary does not move, pressure-volume work is zero, and DeltaU=Q\\Delta U=Q. Over a complete cycle the system returns to its initial state, so DeltaU=0\\Delta U=0; net heat input equals net work output.

Example: Compute the internal-energy change

A system receives 500500 J of heat while expanding and doing 180180 J of work on its surroundings. Find ΔU\Delta U.

Solution

Using ΔU=Q−Wby\Delta U=Q-W_{by}, ΔU=500−180=320\Delta U=500-180=320 J.

Quick check

The first law applies energy conservation to a thermodynamic system, but first choose the system and sign convention. With ΔU=Q+Won\Delta U=Q+W_{on}, compression means Won>0W_{on}>0; if the gas expands and pushes a piston, work done on the system is negative. For a complete cycle, the final internal energy equals the initial value, so ΔU=0\Delta U=0, even though heat and work individually need not vanish. This means net heat received equals net work delivered by the system. During an isochoric process, pressure-volume work is zero, so heat entering the system changes its internal energy. The law does not determine whether a process occurs spontaneously or what efficiency it can achieve; that is the role of the second law. In each calculation, write the sign and units of every energy term before adding them, and do not confuse work by the gas with work on the gas.

For example, if a system absorbs 120 J of heat and expands while doing 45 J of work on the surroundings, then ΔU=120−45=75\Delta U=120-45=75 J. Writing Won=−45W_{on}=-45 J gives the same result. This checks the signs: expansion does not add energy through work done on the system.

A system absorbs 200200 J of heat and expands, doing 5050 J of work on the surroundings. What is ΔU\Delta U?

What is the change in a system's internal energy over a complete cycle?

References

  1. Charles Kittel, Herbert Kroemer (1980). Thermal Physics