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Calculate Water Heating Time: 2000W Heater, 140L Tank, 50 Degrees - Optimal Boiler Settings

spider_net 50991 36
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How long will it take to heat 140 L of water to 50°C with a 2000 W heater, and how should a boiler be controlled for best savings with a programmable socket?

Use the heat formula Q = c·m·ΔT and then t = Q/P; for water, c ≈ 4.19 kJ/kgK and 1 L ≈ 1 kg [#15617665][#15617784] For a 140 L tank heated from 15°C to 50°C, the ideal heating time is about 2.85 hours. If the heater only warms part of the tank, use that mass instead; one reply calculated 70 L from 10°C to 50°C as 11.76 MJ, which gives about 1.63 hours [#15621040] In practice the real time will be longer because of heat losses, cold water inflow, and heater placement, so the result is only approximate [#15617651][#15619566][#15621637] For savings, a fixed timer is mainly useful when you are away and do not want to heat the tank all day; the better solution is a thermostat or temperature sensor, with the boiler set to the lowest acceptable temperature, often around 35–40°C in the thread [#15617627][#15617799][#15617838]
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  • #31 15621040
    idepopizze
    Level 33  
    Posts: 2467
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    @spider_net

    E = Cw * m * (Tk-Tp)

    E - The heat energy required to change the temperature J
    Cw - specific heat of water J / kg * K.
    m - water weight (1l of water = 1kg)
    (Tk-Tp) - temperature difference between the beginning and end of heating

    Tp = 10C - water temperature from the water supply
    Tk = 50C
    m = 70 liters - because the heater is in the half of the tank and heats only what is above it.

    E = 4,200 * 70 * (50-10) = 11,760,000 J

    This is the amount of energy needed to heat 70 liters of water. Of course, system efficiency and losses are neglected.

    Heating time.

    P = E / Vol

    t- time in seconds

    t = E / P
    P = 11,760,000 / 2,000 = 5,880 seconds = 1.63 hours

    it hurt
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  • #32 15621194
    CYRUS2
    Level 43  
    Posts: 17639
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    The power plant is not supplying power to the power heater terminals - only voltage.
    P = U? / R
    5% voltage = 10% calculation error. Just because of the tension.
    Dear physicists!
    It makes no sense to count anything if we do not know the error required.
    Author, tell me what you need this time for. !!!!
    Because we need these calculations for something.
    Yes, you can only count on the theoretical needs of a teacher at school.
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  • #33 15621309
    zen_gac
    Level 10  
    Posts: 45
    Rate: 11
    vodiczka wrote:
    zen_gac wrote:
    I have an 80L boiler without circulation. The heater is centrally located in the center.
    Agreed, but the problem is the location of the heater, not the capacity of the tank or the existence or non-existence of circulation. When the heater is placed in the lower zone of the tank (say 5-10 cm above the bottom), natural circulation is created due to the difference in density of hot and cold water.


    In my opinion, the author of the topic asked a practical, not theoretical question - not taking into account the construction of the boiler. There are boilers with a heater at the bottom? And let's not theorize that you can make one in the garage. Sure and you can, but what boiler does the author of the topic have? Self-made with heating at the bottom, or "normal" production, commercially available.

    I agree with what CYRUS2 wrote in post # 32:
    "It makes no sense to count anything if we don't know the required error.
    Author, tell me what you need this time for. !!!!
    Because we need these calculations for something.
    So you can only count on the theoretical needs of a teacher at school. "
  • #34 15621637
    vodiczka
    Level 43  
    Posts: 30168
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    zen_gac wrote:
    There are boilers with a heater at the bottom?
    There are, I have one hanging vertically, 50l. It is possible that those with higher capacity do not have such a solution.
    zen_gac wrote:
    "It makes no sense to count anything if we don't know the required error.
    In this case it makes sense because without taking into account the heat losses, we only get an approximate result.
    idepopizze wrote:
    P = 11,760,000 / 2,000 = 5,880 seconds = 1.63 hours
    These 3/100 actually make no sense, but you can assume that the heating time will be approximately 2 hours This is enough for the first setting and the average real time will be known after 2-3 heating attempts.
    Once, out of curiosity, I counted the time needed to boil the water in the kettle (assuming 230V voltage, no heat loss, measured water outlet temperature) and compared it with the time measured with my watch. I do not remember exactly but the difference did not exceed 20%
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  • #35 15622006
    rafbid
    Level 33  
    Posts: 2450
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    zen_gac wrote:
    Author, tell me what you need this time for. !!!!
    He wrote in the first post that for the savings that may result from heating the water before using it, it does not have to be warm all day, or it has two tariffs and wants to heat it cheaper.
  • #36 15622067
    retrofood
    VIP Meritorious for electroda.pl  
    Posts: 31317
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    Do you want a PhD in this? Have fun elsewhere, not here. A simple formula, assuming the assumed ideal conditions, and the corrections are introduced by the Author himself, because he knows the losses himself. So much.
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Topic summary

✨ The discussion revolves around calculating the heating time for water in a 140L tank using a 2000W heater to reach a temperature of 50 degrees Celsius. Various participants provided insights into the necessary calculations, emphasizing the importance of initial water temperature, specific heat of water, and potential heat losses. A formula was shared: E = Cw * m * (Tk - Tp), where E is the energy required, Cw is the specific heat of water, m is the mass of water, and Tk and Tp are the final and initial temperatures, respectively. The estimated heating time was discussed, with calculations suggesting approximately 1.63 hours under ideal conditions, though real-world factors such as heat loss and water circulation were noted as significant variables affecting the actual time required. The conversation also touched on the practicality of using programmable sockets for energy savings and the importance of knowing the specific conditions of the water supply.
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FAQ

TL;DR: Heating 140 L by 35 °C with a 2 kW element takes ≈2.9 h under ideal conditions; "It's best to measure" [Elektroda, arecoag, post #15617692] Calculated times vary ±20 % due to losses, voltage and heater position.

Why it matters: Correct run-times cut standby losses and cut your power bill.

Quick Facts

• Specific heat of water: 4.19 kJ kg⁻¹ K⁻¹ [Elektroda, Darom, post #15617665] • Density of water: 1 kg L⁻¹ [Elektroda, Darom, post #15617665] • 140 L × 35 °C rise = 20.5 MJ → 2.85 h at 2 kW (ideal) [Calculation, Darom #15617784] • Typical standing loss for 150 L class-C cylinder: 1.5 kWh day⁻¹ [EU ErP 812/2013] • Average EU household electricity price: €0.30 kWh⁻¹ [Eurostat, 2024]

1. What formula gives water-heating time?

Time (s) = (Specific heat × Mass × Temperature rise) ÷ Heater power. Use 4180 J kg⁻¹ K⁻¹, 1 kg = 1 L, and watts for power [Elektroda, Darom, post #15617784]

2. How long to heat 140 L from 15 °C to 50 °C with a 2 kW rod?

Insert the numbers: 4180 × 140 × 35 = 20.5 MJ. Divide by 2000 W → 10 260 s ≈ 2.85 h [Calculation, Darom #15617784].

3. Why did some users report 4–5 h?

Their calculations included heat losses and colder inlet water (≈5 °C) which raise energy need to ≈29 MJ, stretching time to about 4.1 h [Elektroda, Anonymous, post #15617646]

4. What if the thermostat probe and heater sit halfway up the tank?

Only the upper 70 L heats. Time drops to ≈1.6 h, but the bottom stays cold, so usable hot volume halves [Elektroda, idepopizze, post #15621040]

5. Step-by-step: how do I calculate my own tank?

  1. Measure cold-water temperature.
  2. Decide target temperature.
  3. Apply the formula in Q1 using your volume and element power. Done—compare with a stopwatch to refine losses.

6. How much energy do standby losses burn?

A 150 L class-C cylinder loses about 1.5 kWh every 24 h, worth €0.45 at EU average prices [EU ErP 812/2013; Eurostat, 2024].

9. Should I trust the math or just measure?

Both. Calculations give first estimates; stopwatch tests include real losses. "Measure, don't guess" remains best practice [Elektroda, arecoag, post #15617692]

10. Will a programmable socket save money?

Savings appear only if the tank would otherwise stay hot unused for hours. Below 40 °C the gain is small [Elektroda, vodiczka, post #15617799]

11. What’s an edge case where timing fails?

Heating 140 L when you actually draw 2 L daily wastes ~98 % of energy—automation cannot fix chronic oversizing [Elektroda, Darom, post #15618580]
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