How much energy is used in heating water?

[paulfitz1592]

1. I'm trying to work out how much energy is used in heating 2 cubic metres of water per day from 23C to 65C. Basically trying to put a water purifier in Malawi and run it off solar panels. It currently uses 1.3kWh to run the machine but the initial stage involves pre heating water. In Malawi this won't be needed so I'm trying to work out how much less energy the machine will need to run.

2. I understand that you need 1 calorie to heat 1 ml of water 1C.3. Heating the water a total of 42C.
2 cubic meters of water is 2,000,000 ml.
Therefore the machine uses 84,000,000 calories to heat the water, or so I think.
If I use a calorie to kilowatt/hour as 1.163x10-6.
84,000,000 x 1.163x10-6 = 97.7kWh

This is obviously wrong and I'm unsure where, initially thinking maybe because the machine doesn't heat the water all at once as does it gradually over the course of 24 hours?

Any help would be appreciated. I understand this is probably silly on my part but its been a long time since I done physics.

Thanks

 

[AbhinavJ]

Water has a density 1000 kg/m^3
2m^3 = 2000kg
Q=ms(t2-t1)
=2000*4.156*42 J/Kg
=349188j/kg
=0.09699 kW hours

 

[AbhinavJ]

You I messed up with the units I suppose

 

[paulfitz1592]

You messed up with the units I suppose

Doesn't surprise me! Thanks.

 

[Bystander]

Somebody messed up with units --- care to guess who?

 

[AbhinavJ]

Somebody messed up with units --- care to guess who?

Did I? Lol

 

[Bystander]

You both need to go through it again. One for confidence in calculation rather than intuition, and the other for attention to detail.

 

[AbhinavJ]

You both need to go through it again. One for confidence in calculation rather than intuition, and the other for attention to detail.

So, where am I wrong?don't get it

 

[Bystander]

So, where am I wrong?

=2000*4.156*42 J/Kg

 

[BvU]

Dear Paulfitz, welcome to PF

You correctly calculated the energy needed to heat up 2000 kg of water 42 degrees at 4.1868 kJ/kg/$^\circ$C.

What makes you think that the result is "obviously wrong" ?

It currently uses 1.3kWh to run the machine but the initial stage involves pre heating water

Does the machine use 1.3 kW of power while running, or does it use 1.3 kWh per day ? Where does the energy for the preheating come from ? Could it be that there is a gas bill to pay also ?

 

[paulfitz1592]

Dear Paulfitz, welcome to PF

You correctly calculated the energy needed to heat up 2000 kg of water 42 degrees at 4.1868 kJ/kg/$^\circ$C.

What makes you think that the result is "obviously wrong" ?

Does the machine use 1.3 kW of power while running, or dioes it use 1.3 kWh per day ? Where does the energy for the preheating come from ? Could it be that there is a gas bill to pay also ?

As far as I'm aware it uses 1.3kW whilst running so 1.3kWh continuous. My initial calcs involved saving 97kWh which isn't possible surely as it doesn't require this much energy to fully function? But I think my units are wrong and i should have divided by 1000 for kj/kg instead of kj?

We are planning on using black plastic tubing which in the African sun will heat the water to between 60-70C, I'm just trying to find ways to cut down the amount of solar panels needed to power the machine.

 

[Chestermiller]

You messed up with the units I suppose

Actually, you did. The heat capacity of water is 4.156 kJ/kg-C, not 4.156 J/kg-C.

Chet

 

[AbhinavJ]

You both need to go through it again. One for confidence in calculation rather than intuition, and the other for attention to detail.

Thanks, that couldve been a major issue had such a question come in my exams
That's 4.18 j/g isn't it?

 

[BvU]

1.3 kW is 1.3 kWh/h . If the thing runs 24 hours/day that means it uses 31.2 kWh/day. Still not enough to heat up 2 m³ of water from 23 $^\circ$C to 65 $^\circ$C in one day.

 

[Chestermiller]

Thanks, that couldve been a major issue had such a question come in my exams
That's 4.18 j/g isn't it?

Yes. j/g-C.

 

[BvU]

JA has edited his post #2 which makes the rest of the thread unintellegible. Great.

Let's get back to Paul and help him (her?) out. The 1.3 kW can't be right if the other numers are correct. What's missing ?

 

[Chestermiller]

JA has edited his post #2 which makes the rest of the thread unintellegible. Great.

I'm going to restore the original.

Chet

 

[paulfitz1592]

1.3 kW is 1.3 kWh/h . If the thing runs 24 hours/day that means it uses 31.2 kWh/day. Still not enough to heat up 2 m³ of water from 23 $^\circ$C to 65 $^\circ$C in one day.

'It is electrically driven (rated at 1.3kWh) ' that's the information I have from the company.

 

[BvU]

Any links you might want to share ?

 

[haruspex]

'It is electrically driven (rated at 1.3kWh) ' that's the information I have from the company.

Sadly, it's not unusual to see confusion between kW and kWh in such a place, where you'd think they understood this stuff.

 

[paulfitz1592]

Sadly, it's not unusual to see confusion between kW and kWh in such a place, where you'd think they understood this stuff.

Sorry I'm unsure what you mean by that. Are you saying the machine is powered at 1300 kW and not 1300 kWh?

 

[BvU]

'It is electrically driven (rated at 1.3kWh) ' that's the information I have from the company.

That is very weird. Normally the rating of a motor has the dimension of power, not energy. I don't think I would buy from such a company !

Power = energy / time. Units: Watt (= Joules/s), kW (=kJ/s), etc.

Energy: Units : Joule, kJ, kWh (= 1000 W an hour long, so 3600 kJ or 3.6 M J )

 

[haruspex]

Sorry I'm unsure what you mean by that. Are you saying the machine is powered at 1300 kW and not 1300 kWh?

Yes, except that I think you mean "1.3kW (1300W) and not 1.3kWh". "Powered at 1.3kWh" doesn't mean anything. It's like saying a car's top speed is 100km.
To show how widespread confusion is on this, I was recently sent some usage data from an electricity supplier. The half-hourly data had a column titled "kW", but it turned out that the numbers were actually kWh. That doubled the usage.