Designing an Exam Question: Calculate the Energy Needed to Heat a Concert Hall

Hook: From stage lights to steady-state heat — why this problem matters for exam takers

Struggling to connect abstract heat-transfer equations to real-world exam questions? You’re not alone. Students often freeze when an exam asks for HVAC energy in a building-sized problem because the physics looks messy and the assumptions multiply. This worked exam-style problem bridges that gap: inspired by Mitski’s tour vibes — intimate venues, stage heat, and late-night shows — you’ll learn to model a concert hall’s heating needs, perform a clear calculation, and present a concise answer that examiners can mark easily.

The exam-style question (scenario + tasks)

Context

A touring artist is performing a three-hour show in a rectangular concert hall. The venue manager asks for the HVAC heating energy required to keep the hall at a comfortable 20°C when the outside temperature is 0°C. The problem tests steady-state heat transfer, ventilation heat load, internal heat gains, and basic HVAC system energy accounting.

Given data (treat as exam givens)

• Hall dimensions: length = 40 m, width = 25 m, height = 12 m.
• Occupancy during the show: 1500 people.
• Ventilation requirement: 8 L s^-1 per person (mechanical supply of outside air).
• Internal sensible gains: 80 W per person (sensible heat — dancing/audience activity).
• Stage and house lighting & sound equipment combined sensible heat: 35 kW.
• Envelope U-values: walls U = 0.70 W m^-2 K^-1, roof U = 0.35 W m^-2 K^-1, floor U = 0.40 W m^-2 K^-1.
• Outside temperature = 0°C; required inside temperature = 20°C. (ΔT = 20 K)
• Air properties: density ρ = 1.20 kg m^-3, specific heat c_p = 1005 J kg^-1 K^-1.
• Assume steady-state conditions and that internal gains are entirely sensible. Ignore latent loads and infiltration for the base case.

Tasks

1. Calculate the steady-state heating power (in kW) that the HVAC must supply to maintain 20°C.
2. Calculate the total heating energy required for the 3-hour concert (in kWh).
3. Two system variants: (A) a gas boiler with 90% efficiency; (B) an electric heat pump with an average COP = 3.5. For each variant compute the primary energy input (fuel kWh for boiler, electrical kWh for heat pump) required for the 3-hour event.

Worked solution — step-by-step (showing exam-style clarity)

Step 1 — Volume and geometric areas

Volume V = length × width × height = 40 × 25 × 12 = 12 000 m³.

Perimeter walls area: A_walls = 2 × (40 + 25) × 12 = 2 × 65 × 12 = 1560 m².

Roof area: A_roof = 40 × 25 = 1000 m². Floor area: A_floor = 1000 m².

Step 2 — Conduction (envelope) heat loss

Compute UA for each element (U × area):

• Walls: UA_walls = 0.70 × 1560 = 1092 W K⁻¹.
• Roof: UA_roof = 0.35 × 1000 = 350 W K⁻¹.
• Floor: UA_floor = 0.40 × 1000 = 400 W K⁻¹.

Total UA = 1092 + 350 + 400 = 1842 W K⁻¹.

Conduction loss power P_cond = UA × ΔT = 1842 × 20 = 36 840 W ≈ 36.8 kW.

Step 3 — Ventilation heat load

Ventilation flow per person = 8 L s^-1 = 0.008 m³ s^-1.

Total ventilation flow V̇ = 1500 × 0.008 = 12.0 m³ s⁻¹.

Mass flow ṁ = ρ × V̇ = 1.20 × 12.0 = 14.4 kg s⁻¹.

Ventilation heating power P_vent = ṁ × c_p × ΔT = 14.4 × 1005 × 20 ≈ 14.4 × 20 100 ≈ 289 440 W ≈ 289.4 kW.

Step 4 — Internal sensible gains

People: 80 W per person × 1500 = 120 000 W = 120 kW.

Lighting & equipment: 35 kW.

Total internal sensible gains P_int = 120 + 35 = 155 kW.

Step 5 — Net heating power required

Summing losses and subtracting internal gains:

P_required = P_cond + P_vent − P_int = 36.8 + 289.4 − 155 = 171.2 kW.

Interpretation: The ventilation load dominates (≈ 289 kW) — common in dense-occupancy venues. Internal gains offset a large fraction but not enough to eliminate heating need.

Step 6 — Energy for the 3-hour concert

Energy E = P_required × time = 171.2 kW × 3 h = 513.6 kWh.

Step 7 — System variants: primary energy inputs

Variant A — Gas boiler (efficiency η = 90%): fuel energy required = E / η = 513.6 / 0.90 ≈ 570.7 kWh (fuel).

Variant B — Electric heat pump (COP = 3.5): electrical energy required = E / COP = 513.6 / 3.5 ≈ 146.7 kWh (electric).

Rounded final answers (what to write in your exam)

• Steady-state heating power needed: ≈ 171 kW.
• Total heating energy for a 3-hour show: ≈ 514 kWh.
• Gas boiler (90%): fuel input ≈ 571 kWh.
• Electric heat pump (COP 3.5): electricity ≈ 147 kWh.

Marking scheme (exam-friendly rubric)

Full marks: 20 marks. Mark allocation shows what examiners expect — and the precise steps you should show.

1. Set-up & geometry (3 marks)
   • Correct volume and areas calculated with units (2 marks)
   • Correct identification of ΔT = 20 K (1 mark)
2. Conduction calculation (3 marks)
   • UA for each element shown with units (2 marks)
   • Pcond result with correct multiplication by ΔT (1 mark)
3. Ventilation heat load (4 marks)
   • Ventilation flow per person converted to m³ s⁻¹ and total V̇ (1 mark)
   • Mass flow ṁ and use of cp and ΔT to get Pvent (2 marks)
   • Correct numeric answer with units (1 mark)
4. Internal gains (2 marks)
5. Net heating power (3 marks)
   • Correct algebra: Pcond + Pvent − Pint (2 marks)
   • Final kW with units (1 mark)
6. Energy/time and system variants (3 marks)
   • Energy for 3 h with correct units (1 mark)
   • Correct fuel/electric input for each variant (2 marks)
7. Clarity & assumptions (2 marks)
   • Clear statement of steady-state, ignored infiltration/latent loads (1 mark)
   • Reasonable rounding and units (1 mark)

Exam tips, common pitfalls, and advanced extensions

Exam tips

• Always show units at each step — examiners deduct for missing units.
• Label your assumptions (e.g., steady-state, ignoring latent heat) — this gains marks for reasoning.
• If ventilation numbers are surprising, double-check unit conversions (L s^-1 to m³ s^-1). Humans are error-prone converting 8 L s^-1 × 1500 people.
• Present final answers with sensible significant figures (2 or 3 s.f.).

Common mistakes

• Forgetting to convert L s^-1 to m³ s^-1.
• Subtracting internal gains incorrectly (treat them as negative heat load — they reduce heating need).
• Mixing kW and kWh — be explicit about power (kW) vs energy (kWh).

Advanced exam extension prompts (for higher-mark questions)

• Include latent loads: use standard moisture generation rates per person and compute dehumidification energy.
• Add heat recovery: assume an energy-recovery ventilator (ERV) with 70% sensible effectiveness and recompute P_vent.
• Introduce transient heating: calculate time to reach 20°C from 10°C using the building thermal capacitance (mass × c_p approach).
• Optimize for decarbonization: compare primary energy and CO₂ for gas boiler vs heat pump, using local grid emission intensity (2026 typical grid intensity can vary — state it clearly).

Why this problem is timely in 2026 — trends and exam relevance

Recent industry focus (late 2024–2025 into 2026) has pushed venue operators to reduce HVAC energy and carbon emissions. Two trends make this problem especially relevant:

• Electrification and heat pumps: Many venues are replacing gas boilers with electric heat pumps (higher COPs, lower site CO₂ depending on grid mix). In exams, it’s common to ask students to compare system-level primary energy and emissions.
• Heat recovery and smart control: Energy-recovery ventilators, demand-control ventilation (DCV) and tour-to-tour sustainability targets (artists increasingly require low-carbon logistics) mean ventilation calculations and recovery efficiency matter in homework and real life.

In short, this question mirrors real engineering trade-offs in 2026: ventilation dominates in congested venues, and choosing the right heating technology and recovery strategy makes large differences to energy use.

Practical, actionable takeaways for students and teachers

• Memorize the ventilation heating formula: P = ρ × V̇ × c_p × ΔT — it appears in many exam questions.
• When in doubt, quantify internal gains. They often offset a surprisingly large fraction of the heating load.
• Practice presenting assumptions clearly — that clarity wins marks even when numerical answers are approximate.
• For teachers: convert this base question into quick exam variants by changing occupancy, ventilation rate, or adding heat recovery to test conceptual depth.

Quick formula cheat-sheet: Conductive loss: P = UAΔT. Ventilation: P = ρV̇c_pΔT. Energy: E (kWh) = P (kW) × time (h).

Short worked example variation (show how small changes matter)

If the venue installs an ERV with 70% sensible effectiveness, effective ventilation heat load becomes P_vent,eff = (1 − 0.70) × P_vent = 0.30 × 289.4 kW ≈ 86.8 kW.

Then new P_required = 36.8 + 86.8 − 155 = −31.4 kW (negative means internal gains exceed losses — cooling may be required). Over a 3-hour show, net energy could be near zero or even a cooling requirement — a real risk in modern venues with efficient envelopes and heavy internal loads. This highlights why ventilation strategy and heat recovery are exam-worthy and operationally crucial.

Wrap-up and call-to-action

Designing exam questions that map to real venues sharpens both physics understanding and practical HVAC judgment. This Mitski-inspired concert-hall problem emphasizes why ventilation heat loads dominate and how internal gains and system choice transform energy needs. Practice this style of problem until you can set up the equations instantly and justify each assumption — that clarity is what examiners reward.

Ready to master more venue-scale thermodynamics problems? Download our bundled practice set (includes ERV, transient, and humidity variants), or try the timed mock exam that uses this marking scheme. Sign up for our study planner to get daily worked problems and video walkthroughs aligned to current 2026 HVAC and building-energy trends.

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