# Calculation Electrical Energy

1. Mar 9, 2009

### zebra1707

1. The problem statement, all variables and given/known data

Electrical Immersion heater unit
Needs to heat 10,000 litres of water from 24 Degrees C to 30 Degrees C in 24 hours
Heater is connected to 240 V.

Specific heat of the water 4.19 kJ kg-1 °C-1 and the energy gained by water is calculated from the formula Q=mc(Delta)T.

What is the current requirement of the heater (ignore energy losses from the water while heating?

2. Relevant equations

3. The attempt at a solution

My attempt

Energy gained = (10,000)(4.19)(30-24)
Energy gained = 251,400 J

Next P = w/t
251,400 J/ 86400 sec (24hours)
P = 2.9 Watts

Next I = P/V
I = 2.9 / 240
I = 0.012 A

Can you please confirm my understanding.

Many thanks Petra d.

2. Mar 9, 2009

### LowlyPion

You dropped a kJ there on your way to figuring watts.

3. Mar 9, 2009

### zebra1707

Hi there

Many thanks for the reply. To clarify, are you saying that the terminology is incorrect, or the calculation is incorrect.

Is the end result correct? Thanks Petra d.

4. Mar 9, 2009

### zebra1707

Hi there again

Are you saying that the 251,400 J should have been 251.4 kJ

Then the calculation would be

P= 2.91 x 10-3 watts

then

I = 1.21 x 10-5 A

Cheers Petra d.

5. Mar 9, 2009

### LowlyPion

I'm observing that you said:
Yet in your calculation used 4.19 per liter.

6. Mar 10, 2009

### zebra1707

Hi LP

Im not sure that I understand your last response. I dont think that I have converted the specific heat to litres?

Cheers Petra d.

7. Mar 10, 2009

### LowlyPion

This is not correct.

The specific heat required to raise a liter of water 1 degree is 4190 not 4.19.

8. Mar 10, 2009

### zebra1707

Hi there

m = 10,000 litres of water (should I have converted this to grams also or left as kg?)

c = 4.19 kJ kg-1 Degree C -1 (should I have converted that to grams? then I would get the 4190 as you have susggested).

Delta T = this is okay - this is clear to me.

Im just getting confused with the specific heat part.

Cheers Petra d.

9. Mar 10, 2009

### zebra1707

Hi there

4.19 kilojoule/kilogram/°C = 4190 joule/kilogram/°C

So the original specific heat is in kJ kg C and Im converting it to j kg to then align with the mass of the water 10,000 litres = 10,000kg to then get a J final answer.

Energy gained = (10,000)(4190)(6) = 251400000 J

P = w/t = 251400000/86400 sec = 2909.7222222

I = P/V = 2909.72/240v = 12.12 A

I think im closer?

Cheers Petra d.

10. Mar 10, 2009

### LowlyPion

Specific heat is just a formula.

You have to be careful with units for any equation you use. It would have been helpful if you remembered that a liter of water was a kg. And of course had you not dropped the kJ/kg part of the constant for water.

Usually it is best to convert to SI units in most all events.

Your answer is greater by a factor of 1000. I should hope you are closer.

11. Mar 10, 2009

### zebra1707

Can you let me know if this is now closer or correct?

Cheers Petra d.

12. Mar 10, 2009

### LowlyPion

Yes it looks correct.

Sorry for being less than clear.

13. Mar 10, 2009

### zebra1707

Thank you very much LP, all your assistance has been extremely valuable.

It is greatly appreciated. Cheers Petra d.