---
title: "First law for control mass, cyclic process"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:first-law-control-mass-derive-cyclic"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/first-law-control-mass-derive-cyclic
site: BCT Engineering
---

# First law for control mass, cyclic process

Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics

- Asked in: 2070 Magh, 6 marks
- Video: [First law for a closed system undergoing a cycle and change of state|1st law of thermodynamics | ET](https://www.youtube.com/watch?v=p9Sn3DmYOSI) — ENGINEER'S THOUGHT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/first-law-control-mass-derive-cyclic
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4

## Question

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

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Joule's experiments show:
   A. Adiabatic work depends only on end states, defining property E
   B. Work depends on path
   C. Heat equals work
   D. E is constant
   Answer: A. Gives -W_ad = E2 - E1. (Key point: Joule)

2. With heat exchange:
   A. Q + W = ΔE
   B. Q - W = ΔE
   C. Q = ΔE
   D. W = ΔE
   Answer: B. Q-W path independent. (Key point: Equation)

3. For a cycle:
   A. ∮δQ = 0
   B. ∮δW = 0
   C. E2 = E1 so ∮δQ = ∮δW
   D. ΔE > 0
   Answer: C. Property returns. (Key point: Cycle)

4. Which is correct?
   A. Q is a state function
   B. W is a property
   C. All are properties
   D. E is a state function; Q and W are path functions
   Answer: D. Difference Q - W is path independent. (Key point: Path vs state)

## Close variants set in other exams

- [First law for control mass in cyclic process](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Afirst-law-control-mass-cyclic) — Fundamental of Thermodynamics & Heat Transfer, chapter 4, asked 2×

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