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First law for control mass, cyclic process

Short answer

The first law for a control mass is Q - W = ΔE = ΔU + ΔKE + ΔPE. For a cyclic process, ΔE = 0, so ∮ δQ = ∮ δW, i.e. net heat equals net work.

Derive and explain first law of thermodynamics for a control mass. Also reduce it for a cyclic process.

Derivation of the first law for a control mass

The first law is based on Joule's experiments: when a closed system undergoes an adiabatic process between two states, the work done depends only on the end states. That defines a property, the energy E:

−W_adiabatic = E₂ − E₁

If heat is also exchanged, the energy of the system changes by the net energy transferred:

Q − W = E₂ − E₁ = ΔE = ΔU + ΔKE + ΔPE

In differential form: δQ − δW = dE; in rate form: Q̇ − Ẇ = dE/dt.

Explanation

  • Q is positive when heat is added to the system, W is positive when work is done by the system.
  • E is a property (state function) and Q and W are path functions, but the difference Q − W is path independent.
  • For a stationary system ΔKE = ΔPE = 0, so Q − W = ΔU = m(u₂ − u₁).

Reduction for a cycle

In a cyclic process the system returns to its initial state, so E₂ = E₁ and ΔE = 0:

∮ δQ − ∮ δW = 0 ⇒ ∮ δQ = ∮ δW

The net heat added in a cycle equals the net work done: Q_net = W_net. Example: for a power cycle, Q_H − Q_L = W_net.

✓ Answer checked by Claude Opus 5.5 (high effort)

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