A heat pump promises to heat your home consuming much less than a gas boiler, because it doesn't "burn" energy but moves it: it draws heat from the air or the ground and transfers it indoors, delivering several units of heat for each unit of electricity consumed. But how much do you really save? It depends on the COP (the pump's efficiency), on energy and gas prices and on the home's demand — factors that can change the result a lot. And there is the higher installation cost to recover. The heat pump vs gas boiler savings calculator compares the annual heating cost of the two solutions, gives you the annual saving and calculates in how many years the lower consumption repays the extra cost. So you see, with your numbers, whether and how much it pays to switch to a heat pump.

What the heat pump saving is

The heat pump saving is the difference between what you spend to heat your home with a gas boiler and what you spend with a heat pump, for the same heat produced. The boiler burns gas with an efficiency below 100% (part of the energy is lost); the heat pump uses electricity with a much higher efficiency, the COP, which is typically 3-4: it means that for each kWh of electricity consumed it produces 3-4 kWh of heat. This makes it far more efficient in energy use. The economic saving, however, also depends on the price of energy versus that of gas: electricity generally costs more than gas per kWh, so the consumption advantage must be compared with this price gap. The calculator isolates precisely this comparison — heating cost with one and with the other — which is the most immediate and often decisive component in the choice.

How it is calculated

The annual cost of each system starts from the home's heat demand (the heat needed in a year, in kWh). For the heat pump: demand ÷ COP × energy price. For the boiler: demand ÷ efficiency × gas price. The annual saving is the difference. Example: demand 12,000 kWh; pump with COP 3.5 and energy at €0.25/kWh; boiler with 90% efficiency and gas at €0.10/kWh. Pump cost: 12,000 ÷ 3.5 × 0.25 = 3,429 × 0.25 = €857. Boiler cost: 12,000 ÷ 0.90 × 0.10 = 13,333 × 0.10 = €1,333. Annual saving: 1,333 − 857 = €476. If the pump costs €6,000 more to install, the payback time is 6,000 ÷ 476 ≈ 12.6 years. From then on you save every year.

The COP: the efficiency multiplier

The COP (Coefficient of Performance) is the heart of the heat pump: it indicates how many units of heat it produces for each unit of electricity consumed. A COP of 3.5 means that 1 kWh of electricity generates 3.5 kWh of heat. It is this "multiplication" that makes the heat pump far more efficient than a boiler (which at most returns almost all the energy of the gas, but no more) and even than an electric resistance heater (COP = 1). The higher the COP, the less electricity is needed and the greater the saving. The COP is not constant: it depends on the outdoor temperature and the temperature at which it must supply the heat. In intense cold the COP drops, because the pump struggles more to extract heat from the cold air. This is why, in the calculation, it is best to use the SCOP (seasonal COP), which is the average COP over the whole heating season and represents the real annual efficiency, not that of a single favorable moment.

The home's heat demand

The heat demand is the amount of heat, in kWh per year, needed to heat the home (and possibly for hot water). It depends on the surface, the insulation, the climate and habits. A well-insulated home may require 5,000-10,000 kWh a year; a poorly insulated or large one may exceed 15,000-20,000 kWh. You can estimate it from the gas bills: the cubic meters (or kWh) of gas consumed in a year, multiplied by the boiler efficiency, give the heat actually produced. Demand determines the scale of the saving: the more heat is needed, the larger the pump's absolute saving (and the faster the extra cost is repaid). Very energy-hungry homes get the maximum benefit from switching to a more efficient system; already very frugal homes save little in absolute value, and the return on investment is slower.

The electricity price

The electricity price is, together with the COP, the factor that most determines the saving. Since the pump consumes electricity, a high energy price reduces the advantage, while low tariffs (or self-production from solar) greatly increase it. The key point is the ratio between electricity price and gas price: the pump is worthwhile when the electricity price divided by the COP is lower than the gas price divided by the boiler efficiency. If electricity costs much more than gas, a high COP is needed to compensate; if the price gap is small, even a moderate COP brings a good saving. With solar, the cost of self-consumed electricity can be very low, making the heat pump extremely worthwhile. In the calculation, use the energy price you really pay, considering any fixed charges and self-consumption.

The gas boiler efficiency

The boiler efficiency indicates how much of the gas energy is actually turned into useful heat. A traditional boiler has an efficiency around 85-90%; a modern condensing boiler can reach 95-100% (or a little more, referred to the lower heating value). A higher efficiency means less gas is needed for the same heat, so the boiler costs less to run and the pump's advantage shrinks. In the comparison, use the real efficiency of your boiler (or of the one you compare): an old inefficient boiler makes the pump appear far more worthwhile, a recent condensing boiler reduces the gap. The efficiency enters the calculation by dividing the demand: demand ÷ efficiency gives the gas actually consumed. Together with the gas price, it determines the annual cost of gas heating with which the pump is compared.

Annual cost as the COP varies

Here is how the heat pump's annual cost (and the saving) changes as the COP varies, with demand 12,000 kWh, energy €0.25/kWh, and boiler at €1,333 (90% efficiency, gas €0.10/kWh):

The COP is decisive: going from 2.5 to 4.5 the pump cost almost halves and the saving quintuples. This is why choosing an efficient, well-sized system with a high SCOP is the key to worthwhileness.

The gas price and how to convert it

The gas price should be entered in €/kWh to compare it correctly with electricity. Often, though, gas is billed in cubic meters (m³): to convert, multiply the price per m³ by the calorific value (about 10.5 kWh per m³, depending on the gas type and local conventions), or — more simply — divide the price per m³ by about 10.5 to get the price per kWh. For example, gas at €1.05/m³ equals about €0.10/kWh. Some bills already show the price in €/kWh: use that, including if possible the variable components (commodity, transport) and excluding the fixed charges, which don't depend on consumption. A high gas price brings the pump's worthwhileness closer; very cheap gas pushes it away. Since energy and gas prices vary over time and by country, it is worth redoing the calculation with updated values.

The extra cost and the payback time

The heat pump generally costs more to install than a boiler, especially if it requires system changes (radiators, outdoor unit, connections). This extra cost is what the annual saving must recover. The payback time is the extra cost divided by the annual saving: it indicates in how many years the additional investment "pays off". A short payback (a few years) makes the choice very worthwhile; a long payback must be compared with the useful life of the system (often 15-20 years): if the payback comes earlier, in the following years you save net. Incentives for installation (where present) reduce the extra cost and greatly shorten the payback: in the calculation, enter the extra cost net of incentives. A low extra cost (or zeroed by incentives) combined with a good COP and favorable prices can make the pump worthwhile right away.

Saving by demand and COP

The saving depends on the scale (demand) and the efficiency (COP). Here are some scenarios with energy €0.25/kWh, boiler 90% efficiency and gas €0.10/kWh:

At the same COP, the more heat is needed the larger the saving; at the same demand, a higher COP greatly increases the saving. The best case is a home that consumes a lot combined with an efficient pump.

The role of climate and low temperatures

The climate matters in two opposite ways. On one hand, a cold climate increases the heat demand (more heat is needed), so the pump's absolute saving grows. On the other hand, with very low outdoor temperatures the COP drops, because the pump struggles more to extract heat from the cold air: the real efficiency falls precisely when more heat is needed. Modern pumps work well even at harsh temperatures, but with a lower COP than in mid-seasons. This is why it is important to use the SCOP (average seasonal efficiency) and not the nominal COP measured in favorable conditions. In very cold areas, some pumps integrate a backup electric resistance (COP = 1) on the harshest days, which lowers the average efficiency. Consider your area's climate when estimating the SCOP: a realistic value for your context is essential for a reliable calculation.

Supply temperature: radiators vs underfloor heating

A decisive technical factor for the COP is the supply temperature, that is how hot the water the pump supplies to the system must be. The lower the required temperature, the higher the COP. Underfloor radiant panels, which work at low temperature (30-40 °C), are ideal for the heat pump and allow high COPs. Traditional radiators, designed for water at 60-70 °C, require a greater effort from the pump, lowering the COP and hence the saving. This is why, in homes with radiators, it may pay to combine the pump with system changes (larger radiators, fan coils) or accept a lower COP. This aspect explains why the same pump model can have very different efficiencies in different homes: the calculator uses the COP/SCOP you enter, so reflect in the estimate the type of system (low or high temperature) you have or plan.

Beyond consumption: incentives and other factors

The calculation focuses on the cost of energy and gas, the most immediate item, but the complete picture includes more. Incentives for installation (deductions, grants) can greatly reduce the extra cost, sometimes zeroing it, with a huge impact on the payback. The maintenance of the pump is generally simple, but requires periodic checks; the boiler has maintenance costs and flue inspections. The heat pump often also offers summer cooling (it works as an air conditioner), a benefit not monetized in the calculation but real. Then there is the matter of the meter's electrical power: a pump may require a power increase, with a small additional cost. Finally, switching to electric reduces direct emissions, especially if the energy is renewable. These factors, not included in the comparison of consumption alone, mostly tend to favor the heat pump.

Factors in worthwhileness

Here is a summary of the factors that determine how worthwhile the heat pump is:

The ideal case combines a high COP, favorable prices, significant demand and a low extra cost. When these factors align, the heat pump pays back quickly and saves a lot; when they are unfavorable (low COP, expensive electricity, cheap gas), the advantage can shrink or disappear.

How the calculator works

Enter the annual heat demand (kWh), the pump's COP/SCOP, the energy price (€/kWh), the boiler efficiency (%) and the gas price (€/kWh). Optionally add the pump's installation extra cost. The calculator gives you the annual saving, the pump's annual energy cost, the boiler's annual gas cost, the payback time of the extra cost and an assessment of worthwhileness. In a few seconds you see, with your numbers, how much you save and in how long the heat pump repays its higher installation cost.

The data you need

You need five data (plus an optional one). The annual heat demand in kWh (estimable from gas bills × efficiency). The pump's COP/SCOP: use the realistic seasonal SCOP for your climate and system type, not the optimistic nominal COP. The energy price in €/kWh (considering any solar). The boiler efficiency in % (85-90% traditional, 95-100% condensing). The gas price in €/kWh (convert from €/m³ by dividing by about 10.5). Optional: the installation extra cost net of incentives, to calculate the payback. Realistic data, especially the SCOP and prices, are the key to a reliable result.

Frequently asked questions

How much do you save with a heat pump vs gas?

It depends on COP, prices and demand. In a typical case (12,000 kWh, COP 3.5, energy €0.25/kWh, boiler 90% and gas €0.10/kWh) the saving is about €476 a year. With a higher COP or favorable prices (or solar) the saving grows a lot.

How is the saving calculated?

Pump cost = demand ÷ COP × energy price; boiler cost = demand ÷ efficiency × gas price. The saving is the difference. Example: boiler €1,333 − pump €857 = €476 a year. The calculator makes this comparison and estimates the payback of the extra cost.

What is the COP and why is it so important?

The COP (Coefficient of Performance) is how many units of heat the pump produces for each unit of electricity: a COP of 3.5 means 3.5 kWh of heat per 1 kWh of electricity. The higher it is, the less electricity is needed and the more you save. It is the factor that most determines worthwhileness.

What is the difference between COP and SCOP?

The COP is measured in a specific condition; the SCOP is the seasonal average, that is the real efficiency over the whole heating season. Since the COP drops in cold weather, always use the SCOP (lower and realistic) in the calculation, not the optimistic nominal COP.

How long does it take to repay the heat pump?

By dividing the installation extra cost by the annual saving. With a €6,000 extra cost and €476 saving, the payback is about 12.6 years. Incentives reduce the extra cost and greatly shorten the payback; enter the extra cost net of incentives.

How do I estimate the home's heat demand?

From the gas bills: take the kWh (or m³ × ~10.5) of gas consumed in a year and multiply by the boiler efficiency. You get the heat actually produced, that is the heat demand. An average home often consumes 10,000-15,000 kWh a year.

Why does the system type (radiators vs underfloor) matter?

Because the COP depends on the supply temperature: the lower it is, the higher the COP. Underfloor panels (30-40 °C) allow high COPs; traditional radiators (60-70 °C) lower the COP and the saving. Reflect your system type in the SCOP you enter.

Is the heat pump worthwhile even in the cold?

Yes, but with a lower COP: in intense cold the pump struggles more to extract heat from the air, so efficiency drops precisely when more heat is needed. Modern pumps work well even at low temperatures; use a realistic SCOP for your climate.

How do I convert the gas price from €/m³ to €/kWh?

Divide the price per m³ by about 10.5 (the calorific value of gas per m³). Example: €1.05/m³ ÷ 10.5 ≈ €0.10/kWh. Some bills already show the price in €/kWh: in that case use it directly, including the variable components.

What data do I need?

Five: annual heat demand (kWh), COP/SCOP, energy price (€/kWh), boiler efficiency (%), gas price (€/kWh). Optional: the installation extra cost (net of incentives) for the payback. The realistic SCOP is the most important datum.

Does the calculation include incentives and maintenance?

No, it focuses on the comparison of consumption (energy vs gas). Incentives, maintenance, summer cooling and meter power increase are additional factors that vary by country: incentives in particular reduce the extra cost, so enter it already net.

Is it technical advice?

No. It is a tool to compare heating costs and the payback of the extra cost. The result depends on the estimates you enter (especially the SCOP) and on prices, which vary over time. For sizing and system choice, consult a qualified technician.

Updated July 2026