# More efficient way to heat liquids

If we consider two processes to heat up a bucket of water as follows:-

• Taking one bucket and heating it to our desired temperature for V volume.

• Taking two buckets of half the volume and heating one to higher temp. then mixing the two.

Which of the above requires less electricity using a heater which will always achieve same given temperature in both the cases.

Edit:- Assume ideal conditions,no heat loss, isobaric process ,heating begins at t=0 with heater at temperature higher than desired temperature.

P.S.- my confusion arises due to Newton's law of heating considering which,the rate of heating will be considerably less in second case as the difference between temperature of water and heater decreases.

• When you say "with heater at desired temperature" you mean some temperature higher than the desired final temperature of the water, correct? You are also saying the heater would be set to the same temperature in both scenarios, correct?
– JMac
Dec 11, 2018 at 15:26
• Yes in both the cases. Dec 11, 2018 at 15:31

I believe they would actually require the same amount of energy/electricity, given the assumptions you've provided. That said, your reasoning is essentially sound; but there is a small detail missing in your analysis, which is the amount of electricity required to keep the heating element at the operating temperature.

If we assume no temperature losses, as the bucket of water heats up, the rate of heat transfer between the bucket and the element decreases. This also means that the heating element is losing less heat, and therefore requires less electricity to maintain it's temperature. Because of this, it would actually use the exact same amount of electricity as heating the entire mass to a lower temperature; it would just take a lot longer as the transfer rate slows down. This should make intuitive sense. If we assume there are no energy losses; and the net result is the same, it should require equal amounts of energy. Just the rate changes.

That said, in real life, you would absolutely want to heat the single bucket to a lower temperature. This is because the bucket will also be losing heat to the surroundings. The higher the temperature, the quicker it loses this heat. That is very bad for the warm bucket; because not only does it take longer to reach the desired temperature, but that temperature is higher, so there will be more ambient heat losses occurring as you are trying to reach the higher temperature half-bucket.

The most efficient way to heat anything is using the reverse Carnot cycle. This is basically what a heat pump does.

The thermal energy required in your example is equal, but you're thinking about a real-world application. In that case, you should be more specific about the conditions.

The biggest mistake you make, which is understandable no offence, is that you assume your process takes any significant amount of time. Heat is lost over time, so if you were to act really quick, there is not a difference.

• But the Carnot cycles efficiency is dependent on temperature of hot and cold sink.Moreover we do this process in isobaric conditions. Dec 11, 2018 at 15:19
• first, yes, it does depend on temperature but its efficiency (units of thermal energy transported per unit of work) is always greater than 1. The efficiency of a heat pump is 1 over the efficiency of the Carnot cycle. This is because combining the 2 (so multiplying the efficiency) can only give a maximum efficiency of 1. Secondly, I never said you had to use the water. a heat pump can use it's own gas in a closed loop to transport heat from 1 place to another with the use of work. Dec 11, 2018 at 19:25