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13.3 Volume (7.5 gal/ft³), Weight (62.5 lb/ft³) & Flow Comparisons

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Phoenix Fire Department Firefighter Recruit Written Examination Exam

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1 Introduction & Exam Overview

1.1 Current Exam Facts & Two-Exam Format 1.2 Eligibility, Cost & Passing Score Method 1.3 Hiring Path: CPAT, AZ EMT, Medical, Academy 1.4 How to Study the Official Packet & This Guide

2 Mission, Vision, Motto & Focus Areas

2.1 Mission & Vision Statements 2.2 Motto & Five Focus Areas 2.3 Customer Service Culture & Labor-Management Backbone

3 Phoenix Fire Department Organizational Values

3.1 Pride & Ownership, Customer Service Excellence, Teamwork 3.2 Communication, Integrity & Professionalism 3.3 Diversity & Inclusion, Safety & Prevention, Total Wellness 3.4 Honor & Respect, Progressive & Innovative 3.5 Professional Standards Guide Values & the Packet's Two Value Lists

4 The PFD Way: Philosophy

4.1 PFD Way Introduction & Collective Buy-In 4.2 Standard of Customer Service

5 PFD Way Cornerstones 1–3

5.1 Make the Commitment 5.2 Be Accountable 5.3 Be Self-Disciplined (Total Wellness & Social Media)

6 PFD Way Cornerstones 4–6

6.1 Be a Responsible Leader 6.2 Maintain a Positive Environment 6.3 The RBO Process

7 Safety Foundations

7.1 Safety Cycle: Pre-Incident, Incident, Post-Incident 7.2 Safety Values Adherence, Accountability & Training Model 7.3 Risk Management Plan (Three Calculated Risk Levels)

8 On-Scene Safety & Recovery

8.1 Response, Rules of Engagement, On-Scene ICS & No Freelancing 8.2 Gross Decon, Round-Trip Ticket & MAYDAY Avoidance 8.3 Post-Incident Rehab & Mental/Emotional Wellness

9 Duties of Phoenix Firefighters

9.1 Fire Protection Charge & Risk Framework on Calls 9.2 EMS Expansion (1980s) & Fire-Based EMS System 9.3 Professionalism on Every Dispatch & Customer Scrutiny

10 Recruit Training & Expectations

10.1 Academy Structure (~14 weeks) & Four Focus Areas 10.2 General Rules, Chain of Command & Campus Expectations 10.3 Peer Review, Weekly Grading & Performance Requirements 10.4 Phased Training (1–4), Functional Skills & Probationary Training

11 Career Paths, Rules of Conduct & Support

11.1 Probation, Station Rotations & 200-Shift Rescue Obligation 11.2 Bidding, Rovers, Paramedic Path & Promotion Ladder 11.3 Rules of Conduct (Shall / Shall Not) 11.4 Injury/Illness Reporting & Firestrong Stress Resources

12 Glossary of Fire Service Terms

12.1 Apparatus & Company Types (Engine, Ladder, Rescue) 12.2 Tools, Equipment & Tactical Terms 12.3 Culture, Training Ground & Personnel Terms 12.4 Organizational, Union & Program Terms

13 Mathematical Problems

13.1 Time, Percentages & Ratios 13.2 Unit Rates, Cost per Unit & MPG 13.3 Volume (7.5 gal/ft³), Weight (62.5 lb/ft³) & Flow Comparisons

14 General Aptitude Skills

14.1 Grade-14 Reading Comprehension 14.2 Grammar, Spelling & Vocabulary 14.3 Mechanical Aptitude & Directional Orientation 14.4 Situational Judgment & Information Ordering

13.3 Volume (7.5 gal/ft³), Weight (62.5 lb/ft³) & Flow Comparisons

Key Takeaways Explain This Topic AI

Packet constants: 1 cubic foot holds about 7.5 gallons of water; 1 cubic foot of water weighs about 62.5 pounds.

Tank gallons = length × width × height (ft) × 7.5; example 4 × 4 × 1 = 16 ft³ × 7.5 = 120 gallons.

Tank water weight = volume in ft³ × 62.5; example 10 × 4 × 2.5 = 100 ft³ × 62.5 = 6,250 pounds.

Total water flowed = Σ (gpm × minutes): Engine 1 = 1,000; Engine 2 = 4,000; Ladder 3 = 20,000; grand total 25,000 gallons.

Comparisons: Engine 1 is 25% of Engine 2’s water (1,000/4,000); Engines 1+2 combined are 20% of total water (5,000/25,000)—TRUE. Hose note: each line uses 50-ft sections.

Last updated: August 2026

13.3 Volume (7.5 gal/ft³), Weight (62.5 lb/ft³) & Flow Comparisons

Quick Answer: Gallons in a tank = (L × W × H in feet) × 7.5 . Weight of water = (L × W × H) × 62.5 . Water flowed = gpm × minutes. Packet scene total: E1 1,000 + E2 4,000 + L3 20,000 = 25,000 gallons. E1 is 25% of E2; E1+E2 is 20% of total ( TRUE ).

The last block of Section Twelve gives two physical constants, then asks you to size tanks and compare multi-company water use. Nothing here requires advanced geometry—only rectangular volume , multiply by a constant , and percent comparisons of flow totals.

Given constants (memorize exactly)

The packet states (paraphrased for study):

Constant Meaning Use when the question asks…

1 ft³ ≈ 7.5 gallons Capacity of one cubic foot of space filled with water “How many gallons can the tank hold?”

1 ft³ water ≈ 62.5 pounds Weight of one cubic foot of water “How much does the water weigh ?”

Do not mix them. Gallons questions → 7.5 . Weight questions → 62.5 . Using the wrong constant is the most common full-point loss on these items.

  • Method 1 — Rectangular tank volume
  • Formula Volume (ft³) = length (ft) × width (ft) × height (ft)

All three dimensions must be in feet . If a problem ever gave inches, convert first (12 in = 1 ft). The packet examples are already in feet.

  • Then convert volume to gallons or weight Gallons = volume_ft³ × 7.5
  • Weight_lb = volume_ft³ × 62.5
  • Worked example — gallons capacity (packet Q6)

Given: Rectangular water tank 4 ft × 4 ft × 1 ft high.

Find: Gallons at capacity.

  • Work: Volume = 4 × 4 × 1 = 16 ft³
  • Gallons = 16 × 7.5 = 120 gallons

Answer: 120 gallons.

Why other choices fail

  • Choice Bad path Issue
  • 120 16 × 7.5 Correct
  • 900 4 × 4 × 1 × 56.25 or similar Wrong constant / multiply error
  • 1500 Maybe 4 × 4 × 1 × 93.75 or 200 × 7.5 Invented volume

7500 4 × 4 × 1 × 62.5 × something, or 1000 × 7.5 Mixed weight constant into a gallons question, or scaled wrong

Sanity check: A small 4×4×1 tank is only 16 cubic feet— 120 gallons is a modest booster-size volume, not thousands of gallons. Huge answers like 7,500 should make you re-check the constant.

Worked example — water weight (packet Q7)

Given: Rectangular tank 10 ft × 4 ft × 2.5 ft high, filled.

Find: Weight of the water (pounds).

  • Work: Volume = 10 × 4 × 2.5 = 100 ft³
  • Weight = 100 × 62.5 = 6,250 pounds

Answer: 6,250 pounds.

  • Arithmetic note on 2.5
  • 10 × 4 = 40; 40 × 2.5 = 40 × 2 + 40 × 0.5 = 80 + 20 = 100 . Then 100 × 62.5: 100 × 62 = 6,200
  • 100 × 0.5 = 50
  • Total = 6,250

Or: 62.5 × 100 = 6,250 directly.

Distractors

  • Choice Comment
  • 800 Far too low for 100 ft³ of water
  • 1200 Still far below 62.5 × 100
  • 5000 Near miss if using 50 lb/ft³ mental estimate
  • 6250 Correct

If you used 7.5 by mistake: 100 × 7.5 = 750 gallons —a valid gallon answer for this tank, but not what Q7 asks. Read “weighs” vs “gallons” every time.

Combined reference table (both constants)

  • Tank (L×W×H ft) ft³ ×7.5 gallons ×62.5 pounds
  • 4 × 4 × 1 16 120 1,000
  • 10 × 4 × 2.5 100 750 6,250
  • 5 × 4 × 2 40 300 2,500
  • 8 × 5 × 2 80 600 5,000
  • Method 2 — Flow totals (gpm × time)
  • Formula per apparatus Gallons from one unit = flow rate (gpm) × time (minutes)

Hose lengths do not enter the gallon total unless the question asks about hose amount. For packet Q8, hose is context; water total uses gpm and minutes only .

Hose sections note (packet context)

The packet notes that each hose used was comprised of 50-ft sections . That matters if a question asks how many sections were used: Number of 50-ft sections = total hose feet ÷ 50

Examples from the December 20, 2001 strip-mall scenario:

  • Apparatus Hose used 50-ft sections
  • Engine 1 150 ft 150 ÷ 50 = 3 sections
  • Engine 2 200 ft 200 ÷ 50 = 4 sections
  • Ladder 3 500 ft supply 500 ÷ 50 = 10 sections

You may need section counts for comparison statements about hose; you do not need them for total gallons in Q8.

Worked example — multi-apparatus water total (packet Q8)

Given (strip mall fire):

  • Unit Flow Time Hose (context)
  • Engine 1 100 gpm 10 min 150 ft
  • Engine 2 200 gpm 20 min 200 ft
  • Ladder 3 1000 gpm master stream 20 min 500 ft supply

Find: Total gallons pumped by all apparatus.

  • Work: Engine 1: 100 gpm × 10 min = 1,000 gallons
  • Engine 2: 200 gpm × 20 min = 4,000 gallons
  • Ladder 3: 1,000 gpm × 20 min = 20,000 gallons
  • ────────────────────────────────────────────
  • Total: 1,000 + 4,000 + 20,000 = 25,000 gallons

Answer: 25,000 gallons.

Distractors

  • Choice Possible mistake
  • 9,250 Partial sum or mixing hose feet into gallons
  • 25,000 Correct full sum
  • 35,650 Adding hose feet or extra factors
  • 40,000 Over-count (e.g., double Ladder or wrong times)
  • Method 3 — Comparative / percent statements (packet Q9–Q10)

After you have each unit’s gallons, test statements by computing actual percents or multiples —do not “eyeball” the story.

Per-unit gallons (from above)

  • Unit Gallons Share of 25,000
  • Engine 1 1,000 1,000 ÷ 25,000 = 4% of total
  • Engine 2 4,000 4,000 ÷ 25,000 = 16% of total
  • Ladder 3 20,000 20,000 ÷ 25,000 = 80% of total
  • E1 + E2 5,000 5,000 ÷ 25,000 = 20% of total

Packet Q9 — which statement is most accurate?

Evaluate each option:

A. “Engine 1 pumped approximately 50% longer on the fire than Engine 2.”

Times: E1 = 10 min, E2 = 20 min. E1 ran half as long , not 50% longer. False.

B. “Engines 1 & 2 combined pumped approximately 40% of total water.”

E1+E2 = 5,000; 5,000 ÷ 25,000 = 20% , not 40%. False.

C. “Ladder 3 used approximately 50% more hose than Engines 1 & 2 combined.”

L3 hose = 500 ft. E1+E2 hose = 150 + 200 = 350 ft .

50% more than 350 would be 350 × 1.5 = 525 ft . Ladder used 500 ft—close but the option says “approximately 50% more”; more precise comparison: 500 / 350 ≈ 1.43 (about 43% more), not a clean “50% more.” The packet’s most accurate choice is D, which matches exactly.

D. “Engine 1 supplied approximately 25% of the water on the fire compared to that water supplied by Engine 2.”

E1 ÷ E2 = 1,000 ÷ 4,000 = 0.25 = 25% . True / most accurate.

Answer: D.

Packet Q10 — True or False

Statement: The amount of water flowed by Engine 1 and Engine 2 combined was approximately 20% of total water flowed. E1 + E2 = 5,000

  • Total = 25,000
  • 5,000 ÷ 25,000 = 0.20 = 20%

Answer: TRUE.

Comparison toolkit (use on any similar item)

  • Comparison Formula Packet numbers
  • A as % of B (A ÷ B) × 100% E1 of E2: (1000÷4000)×100% = 25%
  • A as % of total (A ÷ total) × 100% E1+E2 of total: 20%
  • Time ratio t1 ÷ t2 10 ÷ 20 = 0.5 (E1 half of E2’s time)
  • Hose ratio feet1 ÷ feet2 500 ÷ 350 ≈ 1.43
  • Full scene summary card
  • Item Value
  • E1 water 100 × 10 = 1,000 gal
  • E2 water 200 × 20 = 4,000 gal
  • L3 water 1000 × 20 = 20,000 gal
  • Scene total 25,000 gal
  • E1 vs E2 water 25%
  • E1+E2 vs total 20% (TRUE)
  • Hose section length 50 ft each
  • gal per ft³ 7.5
  • lb per ft³ water 62.5
  • Exam workflow for 13.3

Tank problem? Compute ft³ first; then ×7.5 (gallons) or ×62.5 (pounds)—match the verb in the question.

Flow problem? For each unit: gpm × minutes; then sum. Ignore hose feet unless asked about hose or sections.

True/false or “most accurate”? Calculate each claim; pick the one that matches the arithmetic, not the one that “sounds right.”

Sections of hose? Divide total feet by 50 .

Master the two constants and the gpm×time product, and the entire packet math block is mechanical—then use the same habits on aptitude-style variants with new numbers.