# Calculation confirmation needed -- energy stored in heated water

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Hi Guys

I'm new the PF and hope I'm posting I the correct place, apologies to admin if not ;-). I'm working on a project at home and maths is not my strong suit. I have an equation that I need to complete and involves converting thermal heat capacity in water into electric power. From what I could find with the following parameters this is the calculation.

Max temp 90degree C
Min temp 22degree C
Water heat retention factor of 4.19KJ/kg.k
Volume of water is 10kg/10L

To get the power or kj energy stored in the water I used this formula
Q= (4.19kj/kg.k)(10kg)((90)-(22))
Q= 2849.2kj of energy
Now using a converter that amount of energy is equal to 791.44 Watt hours of power
Here is the tricky part does that mean to get electrical power I use ohms law.

So 791.44W / 12v = 65.95A current. Is that correct and is that the same as saying it would take 65.95A of current at 12v to heat 10L of water from 22 degrees to 90 degrees ? Excluding efficiently losses ?

Any help would be great !

## Answers and Replies

nasu
Gold Member
What you have calculated is the energy and not power. The 791 Watt-hour is still energy. You just changed the units from joules to Watt-hour.
This is the energy required to heat that water. The power is the rate of providing that energy. It will depend on how fast do you provide the energy to the water. You cannot calculate it form the given information.

The calculation of current and voltage is then meaningless, for the above reason.

What you have calculated is the energy and not power. The 791 Watt-hour is still energy. You just changed the units from joules to Watt-hour.
This is the energy required to heat that water. The power is the rate of providing that energy. It will depend on how fast do you provide the energy to the water. You cannot calculate it form the given information.

The calculation of current and voltage is then meaningless, for the above reason.
Hi Nasu
Thank you for the feedback
How would I go about calculating this then ? The time required to heat the water is irelevent to me. If it takes 8 hours then It takes 8 hours. Just trying to find the formula to calculate how much power or voltage/ current it takes or toke to heat water from temp x to temp y.

nasu
Gold Member
The power is a rate so it depends on the time. There is no way to calculate power if you don't decide about some time.
I thought I explained this already. The power is not determined by the given data. Can by any value.

Of course, for a realistic situation, you need to take into account the fact that as you heat up the water, some heat is lost by the water to the medium. In order to continue the heating process you need to provide energy at a rate higher that the rate of heat lost to the medium.

Maybe it may help if you describe the actual application. But for a realistic case, there is no trivial answer.
You need some heat transfer parameters.

The problem is I can't give a time as I don't have it. I can for argument sake I pick a time frame of an hour but it's not realistic. the application is as follows . If a solar hot water heat collector heats up 10L of water from x to y in a average day of sun 6-9 hours. What is the equivalent power consumed if the same amount of water was heated by a element and battery

nasu
Gold Member
Even the combination "power consumed" is not very meaningful. You need to understand the difference between energy and power (rate of energy consumption or production). You can heat that water with various levels of power, as long as you make sure you compensate for the energy lost to the medium.
Ideally, if the tank is well insulated, you can use almost zero power. But it will take almost infinite time.

What you ask is similar to the situation when the distance between two cities (A and B) is given and you ask what is the speed required to go from A to B.
I hope you realize for this case that there is no unique answer. You can walk or you can drive or you can crawl, you will eventually get there but it will take different time intervals to do it. Same thing with the power.
You can use a flashlight battery or the output of a power plant. Very different levels of power. It will take a long time in the first case and very short time in the second.

billy_joule
I understand your point and appreciate there won't be a unique answer if no time is allocated but to complete the formula let's say I measured in 60 minutes the water temp changed from 20 to 30 degrees via the solar system and the tank was 100% insulated. What would the required battery capacity need to be to achieve the same change in temperature in same amount of time.

nasu
Gold Member
Now you ask about battery capacity. Before you asked about power. They are not the same thing.
Let's stick with power.
If the time is 1 hour, then in order to deliver 791 watt-hour of energy you need a power of 791 watt.
The formula is quite simple:
Energy=Power x Time (E=P x t)
or
P=E/t

If you have a 12 V battery, you need a current
I=P/V=(791/12) A
This is about 65 A.
A car battery can provide this current for a short time but I don't think it can do it for one hour.
You will need a battery with a capacity of at least 65 Ah (ampere-hour) but this is a necessary but not sufficient condition. You need to check what is the current the battery can sustain for a long, steady functioning.

So you may need to use a higher voltage or allow for longer time.

Nasu.

You just did the exact same calculation as I did in the original post and came to the same answer. 12v 65Ah. So I don't understand what was wrong then in the first place ? I'm just using a battery as an example can be a power supply capable of delivering the same current.

russ_watters
Mentor
You just did the exact same calculation as I did in the original post and came to the same answer. 12v 65Ah. So I don't understand what was wrong then in the first place ?
Nasu happened to pick a value for time where the numerical result looks the same. But the units are different: energy / time = power. Or Watt-hours / hours = Watts.

So if you pick 8 hrs, you'll get 1/8th the power, for example.

nasu
Gold Member
I did not pick up the value of time.
In post number 7 it is mentioned a time of 60 min so I did the calculation for this time.
If you do it for a different time, you get other values.
In the OP calculation the time is not taken into account and the value of energy is treated as a power, which is an error, no matter what the numbers are.

sophiecentaur
Gold Member
2020 Award
In a problem like this is is important to Tame the units. Stick with Energy (Joules), to find how much you need to heat the water and then, when you have that answer, convert that to kWh (1kWh = 3.6MJ). You then may need to fool around with Watts, Volts and Current to get the answer you need, given the supply Volts. Just do things a step at a time and keep your nerve. It may not come naturally but the way to do this sort of thing is to write down appropriate formulae and then put the numbers in. The way you can get each step right.

CWatters
Homework Helper
Gold Member
In many countries electricity consumption is measured and charged for in "kilo Watt Hours" so perhaps there is no need to do any conversion other than to divide by 1000....
791 WH = 0.791kWH
1kWH costs about £0.16 in the UK at the moment so 0.791kWH costs about £0.12.

All of the above is about energy. If you want to convert it to power you do need decide how fast you want the water heated (as others have said).

791WH is equivalent to..

791W for 1 hour
395.5W for 2 hours
197.75W for 4 hours
98.88W for 8 hours
etc
1W for 791 hours.

So 791.44WH / 12v = 65.95A current for one hour. Is that correct and is that the same as saying it would take 65.95A of current at 12v for one hour to heat 10L of water from 22 degrees to 90 degrees ?

With the corrections in green, yes.

At 12V it would take
65.95A for 1 hour or
32.975A for 2 hours etc etc

Excluding efficiently losses ?
If the heating element is immersed in the water then virtually 100% of the electrical energy will end up in the water making the process virtually 100% efficient. The main source of heat loss is likely to be through the walls of the water tank.

Thank you for all the feedback guys. It helps my approach to this calculation is not from a energy generating point of view but after the fact, analyze the amount of power it toke to heat the water from x to y. So I won't have the time. You arrive and see the max temp in the tank and use room temperature as min. How much power / energy is used was used to achieve this tempretures.

Since the tank is heated by a solar water system and the average amount of sun shine per day is 8H a day. Do I just then use the figure of 8H for all calculations ?

CWatters
Homework Helper
Gold Member
If you are trying to estimate the amount or cost of mains electricity needed or saved then stick with the kWH figure.

If you are trying to work out the size of a solar panel (in kW) needed to heat that much water per day then dividing by 8h might be appropriate.

Last edited:
russ_watters
Mentor
Thank you for all the feedback guys. It helps my approach to this calculation is not from a energy generating point of view but after the fact, analyze the amount of power it toke to heat the water from x to y. So I won't have the time. You arrive and see the max temp in the tank and use room temperature as min. How much power / energy is used was used to achieve this tempretures.
It sounds like you are still confusing power and energy. Again, if time is not specified, the power can be pretty much anything (1 watt, 10 watts, 1,000,000 watts, etc). Is that the answer you really want? What does this answer give you/what are you going to do with it?

nasu