# Calculation for energy consumed over a period of 24 hours

I'm currently studying the textbook Fundamentals of Electric Circuits, 7th edition, by Charles Alexander and Matthew Sadiku. Chapter 1.5 Power and Energy gives the following practice problem:

Practice Problem 1.6
A home electric heater draws 12 A when connected to a 115 V outlet. How much energy is consumed by the heater over a period of 24 hours?

The answer is said to be 33.12 k watt-hours.

The chapter says that the energy absorbed or supplied by an element from time $$\t_0\$$ to time $$\t\$$ is $$w = \int_{t_0}^t p \ dt = \int_{t_0}^t \nu i \ dt, \tag{1.9}$$

where $$\ \nu \$$ is the voltage, $$\ i \$$ is the current, and $$\w\$$ is energy in Joules. It then says that 1 Wh = 3,600 J.

So, by my calculation, we have

$$\int_0^{24} 12 \times 115 \ dt = \left[ 1380t \right]_0^{24} = 33120 \text{ J}.$$

But, converted to watt-hours, as in the textbook's answer, this would be $$\ \dfrac{33120}{3600} = 9.2 \text{ Wh}\$$. Am I misunderstanding something? Are my units incorrect? Or is this, perhaps, a textbook error?

• division by 3600 indicated a possible conversion of quantity per second to quantity per hour ... are you somehow confusing time intervals? Nov 15, 2021 at 18:08
• A joule is not a watt hour it is a watt second.
– RoyC
Nov 15, 2021 at 18:08
• is dt in your integration 24 seconds or 24 hours? Nov 15, 2021 at 20:03

Let's do some back-of-the-envelope stuff before we start losing decimal places.

The heater uses 12 A at 115 V, that's 1380 W, or just over 1 kW. So in one hour, it uses just over 1 kWh. In 24 hours, it must use a bit more than 24 kWh. To do it exactly, in 24 hours, it uses 1.38 * 24 = 33.12 kWh, as in the book.

So where's your mistake? A watt is a Joule per second, not hour. Multiply equation (1.9) by 3600 to get from hours to seconds, and you'll get the right answer.

• I don't understand. If the textbook says that 1 Wh = 3,600 J, and we have that $w = 33120 \text{ J}$, then don't we just divide 33120 by 3600 to get the amount in Wh? Nov 15, 2021 at 18:15
• You’re integrating over 24 seconds not 24 hours, so your joule figure is 3600 times too small
– Frog
Nov 15, 2021 at 18:20
• @Frog Ohh, yes, I see what you mean. The limits of integration are indeed in units of seconds – not hours – so computing the upper limit should be $1380 \times 24 \times 3600$, so we then get $\dfrac{1380 \times 24 \times 3600}{3600} = 33120 \ \text{Wh}$, as required. Thanks for the clarification. Nov 15, 2021 at 18:25
• It goes without saying that unit consistency is very important. Keep in mind the the Mars Climate Observer was loss back in 1993 because of a navigation error caused by a failure to translate English units to metric. Nov 15, 2021 at 18:29

Am I misunderstanding something?

Yes. The result of the integration from 0 to 24 hours is the total energy in watt-hours, not in joules. You show 33120 J, where it actually is 33120 Wh. Multiply this by 3600 to get 119.232 MWh, or 119.232 MJ.

If you want the final result in joules rather than kWh, a more direct method is to integrate in seconds rather than hours. If you integrate from 0 to 86400, the result in in joules (or watt-seconds).

Physical units means a lot , always keep them in mind to avoid confusion. I’ll give you another solution.

P = VI
E = Pt ==> E = VQ
Q = It = 12 coulombs (amperes × s)
through 24 hours is 1 × 60 × 60 × 24 = 86400 s.

So Q through the day is 12 × 86400= 1,036,800 coulomb (C)
so E = 115 × 1,036,800 = 119,232,000 joules (J).

So as you can see, when confused stick to basics.

• I've never heard of a Qolon. Doesn't turn up any useful internet searches either. ----Wait, do you mean coulombs? Nov 15, 2021 at 19:54
• @Aaron my bad, I edited the answer Nov 15, 2021 at 20:07