Heating Time Calculations – 200cm vs 225cm Hot Tubs with external heater

Assumptions used:

• Water specific heat: 4.186 kJ/kg·°C
• 1L ≈ 1kg
• Heater power: 20 kW and 30kW
• Ideal efficiency (real-world will be slightly longer due to heat loss)

200 cm Round/ 202cm square Hot Tub – 1220 L

Scenario 1: 10°C → 37°C (ΔT = 27°C)

Energy:
Q = 1220 × 4.186 × 27 = 137,833 kJ = 38.29 kWh

• 20 kW heater:
  • Ideal: 1.91 hours
  • Real (winter): 2.6 – 3.0 hours
• 30 kW heater:
  • Ideal: 1.28 hours
  • Real (winter): 1.7 – 2.0 hours

Scenario 2: 4°C → 37°C (ΔT = 33°C)

Energy:
Q = 1220 × 4.186 × 33 = 168,548 kJ = 46.82 kWh

• 20 kW heater:
  • Ideal: 2.34 hours
  • Real: 3.2 – 3.6 hours
• 30 kW heater:
  • Ideal: 1.56 hours
  • Real: 2.1 – 2.4 hours

Scenario 3: 4°C water + (-10°C outside) → 37°C

• 20 kW: 4.0 – 4.8 hours
• 30 kW: 2.7 – 3.3 hours

225 cm Round Hot Tub/ 220cm square – 1660 L

Scenario 1: 10°C → 37°C (ΔT = 27°C)

Energy:
Q = 1660 × 4.186 × 27 = 187,427 kJ = 52.06 kWh

• 20 kW heater:
  • Ideal: 2.60 hours
  • Real: 3.5 – 4.0 hours
• 30 kW heater:
  • Ideal: 1.74 hours
  • Real: 2.3 – 2.7 hours

Scenario 2: 4°C → 37°C (ΔT = 33°C)

Energy:
Q = 1660 × 4.186 × 33 = 228,911 kJ = 63.59 kWh

• 20 kW heater:
  • Ideal: 3.18 hours
  • Real: 4.3 – 4.9 hours
• 30 kW heater:
  • Ideal: 2.12 hours
  • Real: 2.9 – 3.3 hours

Scenario 3: 4°C water + (-10°C outside)

• 20 kW: 5.0 – 6.0 hours
• 30 kW: 3.5 – 4.2 hours

 Core formula

$$Q=m\cdot c\cdot \mathrm{\Delta }T$$

Where:

• Q = required energy (kJ)
• m = mass of water (kg) → ~liters
• c = specific heat capacity of water = 4.186 kJ/kg·°C
• ΔT = temperature difference (°C)

Convert energy to kWh

$$1\text{ kWh}=3600\text{ kJ}$$$$Q_{kWh}=\frac{m\cdot 4.186\cdot \mathrm{\Delta }T}{3600}$$

Heating time

$$t=\frac{Q_{kWh}}{P}$$

Where:

• t = heating time (hours)
• P = heater power (kW), in this case 20 and 30 kW