Quick Answer
The number of kilowatt hours (kWh) needed to heat water in a water heater by 25°C depends on the volume of water being heated and the efficiency of the water heater. The calculation is:
kWh = Volume of water (gallons) x Density of water (lbs/gallon) x Specific heat of water (Btu/lb°F) x Temperature change (°F) / Efficiency of heater x Conversion factor (kWh/Btu)
For a typical 50 gallon residential water heater being heated from 20°C to 45°C (68°F to 113°F) with an efficiency of 0.8, the calculation would be:
50 gallons x 8.34 lbs/gallon x 1 Btu/lb°F x (113°F – 68°F) / 0.8 efficiency x 0.000293 kWh/Btu = 7.84 kWh
So for a typical residential water heater, around 8 kWh would be required to heat the 50 gallons of water by 25°C or 45°F. The actual energy required depends on the specific details like water volume, initial and final temperatures, and heater efficiency.
Heating water accounts for a significant portion of residential energy usage. Water heaters are present in almost every home and need to heat water for uses like bathing, washing dishes, laundry, and more. When sizing a water heater and estimating its energy requirements, an important calculation is determining the energy needed to heat the water by a certain temperature increase.
This calculation allows homeowners and plumbers to properly size a water heater for the hot water needs of a home. It also allows calculation of the electrical or gas energy that will be consumed by the appliance. By understanding the energy usage, homeowners can choose the most efficient water heater for their needs and make informed decisions about operating costs.
In this article, we will walk through the step-by-step calculation to determine how many kilowatt hours (kWh) of electrical energy are needed to heat water in a typical residential water heater by 25° Celsius.
The Formula
The basic formula to determine the energy required to heat water is:
Energy (kWh) = Mass of water (kg) x Specific heat of water (J/kg°C) x Temperature change (°C) / Efficiency of heater x Conversion factor
Breaking this down:
– Mass of water – The mass of water being heated, in kilograms. This is found by multiplying the volume of water (liters or gallons) by the density of water (1 kg/L or 8.34 lbs/gallon).
– Specific heat of water – The amount of energy required to raise one kg of water by 1°C. For water this is 4,186 J/kg°C.
– Temperature change – The number of degrees Celsius the water will be heated.
– Efficiency – The efficiency rating of the water heater (such as 0.8 or 80%).
– Conversion factor – To convert the energy into kilowatt hours, we multiply by 0.000277 kWh/Joule.
Let’s go through an example calculation step-by-step:
Example Calculation
Say we have a typical 120 gallon residential water heater, and we want to know the energy required to heat the full volume from 15°C to 40°C (59°F to 104°F). The heater efficiency rating is 0.8 or 80%.
Step 1. Determine the volume and mass of water
Volume of water = 120 gallons
Density of water = 8.34 lbs/gallon
1 kg = 2.2 lbs
Mass of water = Volume x Density
= 120 gallons x (8.34 lbs/gallon) x (1 kg/2.2 lbs)
= 450 kg
Step 2. Determine the temperature change
Initial temperature = 15°C
Final temperature = 40°C
Temperature change = Final – Initial
= 40°C – 15°C = 25°C
Step 3. Apply the formula
Energy (kWh) = Mass x Specific heat x ΔT / Efficiency / Conversion factor
Energy = 450 kg x 4,186 J/kg°C x 25°C / 0.8 / 0.000277 kWh/J
= 54 kWh
Therefore, to heat 120 gallons of water from 15°C to 40°C, with an 80% efficient water heater, requires 54 kWh of energy.
Variables that Affect Energy Use
From this example, we can see that the major variables that affect the energy required to heat water are:
– Volume of water – The greater the volume of water being heated, the more energy required.
– Temperature difference – The larger the temperature change, the more energy needed.
– Efficiency of the heater – More efficient heaters require less energy input.
– Initial water temperature – The lower the starting temperature, the more energy required.
Other factors like insulation, water chemistry, and distribution piping can also influence efficiency and energy use.
Volume of Water
The volume of water being heated is directly proportional to energy use. Doubling the volume of water would approximately double the energy required. In our example, heating 60 gallons of water instead of 120 gallons would cut the energy required in half, to around 27 kWh.
When sizing a water heater, the first step is determining the peak hot water requirements for the household. This establishes the minimum tank size and heating capacity needed. A properly sized tank will minimize energy waste from cycling on and off.
Temperature Difference
The greater the temperature rise needed, the more energy input necessary. This is why water heaters have adjustable thermostats, to allow homeowners to control the tank temperature.
In some cases, turning down the temperature setting by even 20°F can save a significant amount of energy. This needs to be balanced against having adequate hot water capacity.
Heater Efficiency
Water heater efficiency has dramatically improved over the past decades. Older heaters with efficiencies around 0.5 have been replaced by new models with efficiencies of 0.8-0.95.
Choosing a higher efficiency model can lead to energy savings of 15-30%. The energy savings offset the higher initial costs over the 10-15 year lifespan of a heater.
Electric heat pumps are the most efficient residential water heaters, capable of efficiencies over 3.0. However, the upfront cost is higher.
Initial Water Temperature
The energy content of water rises linearly with increasing temperature. The temperature that water enters the tank from the mains affects the energy input needed.
In our example, we assumed initial temperature of 15°C (59°F). During winter months, incoming water could be closer to 5°C (41°F), increasing energy needs. The opposite occurs in summer.
Knowing the range of inlet water temperatures through the year allows better modelling of overall energy requirements.
Comparing Different Water Heaters
We can use this calculation to compare estimated energy use between different water heater types:
| Heater Type | Volume | Efficiency | Energy to heat 120 gal from 15°C to 40°C |
| Standard electric | 120 gal | 0.8 | 54 kWh |
| High efficiency gas | 120 gal | 0.9 | 48 kWh |
| Electric heat pump | 120 gal | 2.0 | 27 kWh |
Assuming typical hot water usage of around 40-50 gallons per day, an electric heat pump could reduce the daily energy for water heating by 50% or more compared to a standard electric heater.
Over 10-15 years, the energy cost savings of a heat pump water heater can significantly outweigh the higher upfront costs. Homeowners should weigh this tradeoff when choosing a new water heater.
Estimating Seasonal Variation
While we used a simple example with a fixed initial temperature, in reality inlet water temperature fluctuates seasonally. This affects the energy input needed.
To account for this, we can estimate monthly energy requirements based on average inlet temperatures:
| Month | Avg. Inlet Temp °C | Energy for 120 gal (kWh) |
| January | 5 | 60 |
| April | 10 | 57 |
| July | 20 | 51 |
| October | 15 | 54 |
This shows how a 20°C swing in inlet temperature from summer to winter results in around a 15% change in energy requirements for heating.
Knowing these seasonal factors allows homeowners to understand how their utility bills will fluctuate during the year based on water heating loads.
Considering Thermal Losses
An additional factor that can be included is standing losses from the tank over time. As hot water sits in the tank, a portion of the heat will be lost to the surroundings.
Well insulated tanks can reduce this loss, but it should be accounted for in overall energy budgeting.
A rough estimate is that a standard electric water heater may lose 4-6 kWh per day from standing losses in a typical home. This could increase in cold climates or poorly insulated tanks.
Accounting for standing losses, the total daily energy budget would be:
– Heating water use: 27 – 54 kWh
– Standing losses: 4 – 6 kWh
– Total daily energy: 31 – 60 kWh
The standing losses can significantly add to the energy consumption over time, providing further incentive to choose well insulated, high efficiency models.
Converting Between Electrical and Heat Energy Units
When doing these calculations:
– 1 kWh = 3,600,000 Joules
– 1 therm = 105,500,000 Joules
– 1 kWh = 3,412 Btu
Some key conversions:
– kWh x 3,412 = Btu
– Btu x 0.000293 = kWh
Knowing the conversions allows you to work in convenient units tailored to the water heater energy source.
For natural gas water heaters, it is common to see energy usage rated in therms. To convert kWh to therms, divide by 29.3.
Electrical energy can be converted to equivalent fuel energy by multiplying kWh by 3,412 to get Btu rating. This allows comparison of electrical and gas water heater energy consumption.
Accounting for Distribution Losses
One other factor that can influence overall efficiency is distribution losses as hot water flows through pipes to fixtures. Long runs of pipe and recirculation loops can lead to thermal losses and increased energy usage.
Proper insulation of hot water lines helps minimize this. Clustering bathrooms and the kitchen can also shorten the required pipe runs.
Recirculating pump systems improve convenience but use energy to constantly loop hot water. Newer on-demand and smart recirculation controls can help save energy here.
Estimating distribution losses can add 5-15% to the calculated energy usage, depending on the home layout. While complex to model exactly, this loss should be considered in energy use projections.
Key Takeaways
The major points from this discussion on calculating water heating energy requirements:
– Energy use rises with volume of water, temperature rise needed, and lower efficiency.
– Proper sizing of the tank and selecting an efficient heater are key.
– Seasonal inlet temperature swings affect energy needs.
– Standing losses from hot tanks add significantly to energy use.
– Converting between units like kWh, Btu and therms is important.
– Distribution losses from pipes and recirculation can also impact overall efficiency.
Understanding these parameters allows accurate estimation of the energy consumption for residential water heating. This helps homeowners choose the right water heater technology and features to maximize efficiency for their budget and needs.
Proper sizing, insulation, and an efficient heater can lead to significant energy savings on this major home energy user over time. With rising energy costs, informed hot water system decisions are key.
Conclusion
Estimating the energy required to heat water in a home water heater is an important calculation for both homeowners and professionals. By walking through the key parameters step-by-step, we can derive the standard formula:
Energy (kWh) = Volume (gallons) x Density (lbs/gallon) x Specific heat (Btu/lb°F) x Temperature change (°F)
Dividing by the heater efficiency rating and converting units gives the total electrical or fuel energy required.
Properly sizing the tank volume, maximizing insulation, choosing an efficient heater, and minimizing distribution losses are all ways to optimize a hot water system for energy efficiency.
Accounting for the impacts of seasonal temperature swings and tank standing losses provides a more complete picture of the total energy consumed.
Armed with this information, homeowners can make informed decisions when selecting water heating equipment and operating settings to balance performance and energy use for their household needs. Efficient hot water systems are a key factor in reducing home energy costs and environmental impacts.
