Return to Answer

Stovetop vs water heater efficiency

A gas stove-top is about 44% efficient (source), and a modern natural gas water heater with a tank is up to 67% efficient (source (pdf)).

Basic formula

From this answer on Chemistry.SE, the energy needed to heat a liquid from one temperature to another is:

Q = (mass)*(specific heat)*(change in temperature)

For water, the specific heat is simply 1 kcal/(kg °C), so we can leave this out. The mass of water is 1 kg per L.

Total energy needed

Applying the formula:

Q = (5kg)*(100°C - 8°C) = 460kcal

Energy needed from each source

Starting with cold water in the pipes, volume is length * π * radius² = approximately 1.5L:

• In the pot we'll need to heat this 1.5L from 8°C to 100°C = 138kcal
• The remaining 3.5L is heated from 8°C to 49°C in the water heater = 143.5kcal...
• ...and then the rest of the way to 100°C in the pot: 178.5kcal

If the water at the tap is already hot, the breakdown is:

• Heat from 8°C to 49°C in the water heater = 205kcal
• And from 49°C to 100°C in the pot = 255kcal

Final comparison

• Stovetop only: 460kcal / 44% = 1045kcal
• Using the water heater, starting with cold water at the tap:
• (138+178.5)/44% + 143.5/67% = 934kcal
• Using the water heater, starting with hot water at the tap:
• 255/44% + 205/67% = 886kcal

So using hot water from the tap will save anywhere between 11% and 15% compared to using just cold water.

Stovetop vs water heater efficiency

A gas stove-top is about 44% efficient (source), and a modern natural gas water heater with a tank is up to 67% efficient (source (pdf)).

Basic formula

From this answer on Chemistry.SE, the energy needed to heat a liquid from one temperature to another is:

Q = (mass)*(specific heat)*(change in temperature)

For water, the specific heat is simply 1 kcal/(kg °C), so we can leave this out. The mass of water is 1 kg per L.

Total energy needed

Applying the formula:

Q = (5kg)*(100°C - 8°C) = 460kcal

Energy needed from each source

Starting with cold water in the pipes, volume is length * π * radius² = approximately 1.5L:

• In the pot we'll need to heat this 1.5L from 8°C to 100°C = 138kcal
• The remaining 3.5L is heated from 8°C to 49°C in the water heater = 143.5kcal...
• ...and then the rest of the way to 100°C in the pot: 178.5kcal

If the water at the tap is already hot, the breakdown is:

• Heat from 8°C to 49°C in the water heater = 205kcal
• And from 49°C to 100°C in the pot = 255kcal

Final comparison

• Stovetop only: 460kcal / 44% = 1045kcal
• Using the water heater, starting with cold water at the tap:
• (138+178.5)/44% + 143.5/67% = 934kcal
• Using the water heater, starting with hot water at the tap:
• 255/44% + 205/67% = 886kcal

So using hot water from the tap will save anywhere between 11% and 15% compared to using just cold water.

Basic formula

From an answer on Chemistry.SE, this is the formula for energy needed to heat a liquid from one temperature to another:

Q = (mass)*(specific heat)*(change in temperature)

For water, the specific heat is simply 1 kcal/(kg °C), so we can leave this out. The mass of water is 1 kg per L.

Total energy needed

Applying the formula:

Q = (5kg)*(100°C - 8°C) = 460kcal

Energy to preheat

Starting with cold water in the pipes, volume is length * π * radius² = approximately 1.5L:

• In the pot we'll need to heat this 1.5L from 8°C to 100°C = 138kcal
• The remaining 3.5L is heated from 8°C to 49°C in the water heater = 143.5kcal...
• ...and then the rest of the way to 100°C in the pot: 178.5kcal

If the water at the tap is already hot, the breakdown is:

• Heat from 8°C to 49°C in the water heater = 205kcal
• And from 49°C to 100°C in the pot = 255kcal

Stovetop vs water heater efficiency

A gas stove-top is about 44% efficient (source), and a modern natural gas water heater with a tank is at best 67% efficient (source (pdf)).

Final comparison

• Stovetop only: 460kcal / 44% = 1045kcal
• Using the water heater, starting with cold water at the tap:
• (138+178.5)/44% + 143.5/67% = 934kcal
• Using the water heater, starting with hot water at the tap:
• 255/44% + 205/67% = 886kcal

So using hot water from the tap will save anywhere between 11% and 15% compared to using just cold water.

Stovetop vs water heater efficiency

A gas stove-top is about 44% efficient (source), and a modern natural gas water heater with a tank is up to 67% efficient (source (pdf)).

Basic formula

From this answer on Chemistry.SE, the energy needed to heat a liquid from one temperature to another is:

Q = (mass)*(specific heat)*(change in temperature)

For water, the specific heat is simply 1 kcal/(kg °C), so we can leave this out. The mass of water is 1 kg per L.

Total energy needed

Applying the formula:

Q = (5kg)*(100°C - 8°C) = 460kcal

Energy needed from each source

Starting with cold water in the pipes, volume is length * π * radius² = approximately 1.5L:

• In the pot we'll need to heat this 1.5L from 8°C to 100°C = 138kcal
• The remaining 3.5L is heated from 8°C to 49°C in the water heater = 143.5kcal...
• ...and then the rest of the way to 100°C in the pot: 178.5kcal

If the water at the tap is already hot, the breakdown is:

• Heat from 8°C to 49°C in the water heater = 205kcal
• And from 49°C to 100°C in the pot = 255kcal

Final comparison

• Stovetop only: 460kcal / 44% = 1045kcal
• Using the water heater, starting with cold water at the tap:
• (138+178.5)/44% + 143.5/67% = 934kcal
• Using the water heater, starting with hot water at the tap:
• 255/44% + 205/67% = 886kcal

So using hot water from the tap will save anywhere between 11% and 15% compared to using just cold water.

TL;DR: Both methods are about the same, because most of the energy is needed to overcome the "enthalpy of vaporization:" the last push required to go from 100degC to boiling.

From an answer on Chemistry.SE, this is the formula for energy needed to boil a liquid:

Q = (mass)*(specific heat)*(change in temperature) + (mass)*(enthalpy of vaporization)

For water:

• specific heat = 1 kcal/(kg degC)
• enthalpy of vaporization = 540 kcal/kg

Q_heating = (5kg)*(1 kcal/kg/degC)*(100degC - 8degC) = 460kcal Q_boiling = (5kg)*(540kcal/kg) = 2700kcal Q_total = 460kcal + 2700kcal = 3160kcal

From this result it's obvious that regardless of starting temperature, most of the energy is needed to overcome the enthalpy of vaporization, which can only be done by the stove.

But all the same, let's assume I have perfectly insulated plumbing and see if the water heater can help a bit:

Q_preheat = (5kg)*(1 kcal/kg/degC)*(49degC - 8degC) = 205kcal

Thus the remaining energy the stovetop must add is:

Q_remaining = 3160 - 205 = 2955kcal

Using only the stovetop:

Q_stovetop = 3160kcal / 44% = 7182kcal

Preheating with the water heater:

Q_preheat = 205kcal / 67% = 306kcal Q_remaining = 2955kcal / 44% = 6716kcal Q_combined = 306 + 6716 = 7022kcal

So in the end, I've only saved about 160 kcal (186 watt-hours), or 2% of the total energy needed. Given @Fred's point about the possibility of leaching from the water heater, it's probably not worth it.

From an answer on Chemistry.SE, this is the formula for energy needed to heat a liquid from one temperature to another:

Q = (mass)*(specific heat)*(change in temperature)

For water, the specific heat is simply 1 kcal/(kg °C), so we can leave this out. The mass of water is 1 kg per L.

Q = (5kg)*(100°C - 8°C) = 460kcal

Starting with cold water in the pipes, volume is length * π * radius² = approximately 1.5L:

• In the pot we'll need to heat this 1.5L from 8°C to 100°C = 138kcal
• The remaining 3.5L is heated from 8°C to 49°C in the water heater = 143.5kcal...
• ...and then the rest of the way to 100°C in the pot: 178.5kcal

If the water at the tap is already hot, the breakdown is:

• Heat from 8°C to 49°C in the water heater = 205kcal
• And from 49°C to 100°C in the pot = 255kcal

• Stovetop only: 460kcal / 44% = 1045kcal
• Using the water heater, starting with cold water at the tap:
• (138+178.5)/44% + 143.5/67% = 934kcal
• Using the water heater, starting with hot water at the tap:
• 255/44% + 205/67% = 886kcal

So using hot water from the tap will save anywhere between 11% and 15% compared to using just cold water.