---
title: "First law for control mass; cycle; power and refrigeration"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:first-law-control-mass-cycle"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/first-law-control-mass-cycle
site: BCT Engineering
---

# First law for control mass; cycle; power and refrigeration

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

- Asked in: 2073 Bhadra, 6 marks
- Video: [The First Law of Thermodynamics: Internal Energy, Heat, and Work](https://www.youtube.com/watch?v=O7HwhkYt6YU) — Professor Dave Explains
- 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-cycle
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4

## Question

Write down expression for first law of thermodynamics for a control mass. Reduce it for a cycle process and write down statement of first law of thermodynamics for power and refrigeration cycles.

## Short answer

For a control mass, Q - W = ΔU + ΔKE + ΔPE. In a cycle ΔE = 0, so ∮δQ = ∮δW (net heat equals net work). A power cycle gives W_net = Q_H - Q_L, η = W_net/Q_H; a refrigeration cycle needs W_in = Q_H - Q_L, with COP_R = Q_L/W_in.

## Answer

#### First law for a control mass
For a closed system (control mass), the net energy transferred as heat and work equals the change in its total energy:

**Q − W = ΔE = ΔU + ΔKE + ΔPE**

In rate form: dE/dt = Q̇ − Ẇ. For a stationary system, Q − W = ΔU. Sign convention: Q into system +ve, W by system +ve.

#### For a cycle
In a cycle the system returns to its initial state, so ΔE = 0 (E is a property):

**∮ δQ = ∮ δW**  or  Q_net = W_net

#### Statement for power and refrigeration cycles
- **Power cycle:** the net work output equals the net heat input: W_net = Q_H − Q_L > 0, where Q_H is heat received from the high-temperature source and Q_L is heat rejected to the sink. Efficiency η = W_net/Q_H.
- **Refrigeration (heat pump) cycle:** the work input equals the net heat rejected: W_in = Q_H − Q_L, where Q_L is heat removed from the cold space and Q_H is heat rejected to the hot surroundings. COP_R = Q_L/W_in, COP_HP = Q_H/W_in.

In words: for a closed system undergoing a cycle, the algebraic sum of heat transfers equals the algebraic sum of work transfers. (Energy cannot be created or destroyed.)

## Check yourself

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

1. For a control mass the first law is:
   A. Q - W = ΔU + ΔKE + ΔPE
   B. Q + W = ΔU
   C. Q - W = ΔH
   D. Q = W always
   Answer: A. Net energy supplied equals change in total energy. (Key point: First law)

2. In this convention:
   A. Q out and W on are positive
   B. Q into the system and W by the system are positive
   C. Both Q and W are positive into the system
   D. Both positive out of the system
   Answer: B. This is the standard convention used. (Key point: Sign convention)

3. For a cycle ΔE = 0 because:
   A. Q is zero
   B. W is zero
   C. E is a property and the system returns to its initial state
   D. The system is isolated
   Answer: C. This gives Q_net = W_net. (Key point: Cycle)

4. For a refrigeration cycle:
   A. W_in = Q_L - Q_H
   B. W_in = Q_L
   C. W_in = Q_H + Q_L
   D. W_in = Q_H - Q_L
   Answer: D. Work input equals net heat rejected. (Key point: Refrigeration cycle)

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