# Confusion about calculating energy output.

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1. Oct 2, 2015

### srarahcha

The problem statement, all variables and given/known data
I have a question that involves comparing the energy output of a laser to the energy required to make tea. I've already calculated the energy to make tea and I know what to do to find how long it takes to make the tea but I'm struggling with the energy output part.

It says: The Nova laser at Lawrence Livermore National Laboratory in California is used in studies of initiating controlled nuclear fusion . It can deliver a power of 1.60 × 1013 W over a time interval of 2.50 ns.

The attempt at a solution
So initially I multiplied the W by the ns (converted to s) to find the Watts over an interval of 1 second but idk if that worked..... then I divided it instead but I don't think that's right either.... Now i'm going to try just substituting the nanoseconds and the watts into an equation to find the time and see if it works! But i'm still quite confused.

2. Oct 2, 2015

### Bystander

What numbers have you got from your calculation?

3. Oct 2, 2015

### Staff: Mentor

Analyze the units involved. What's the definition of a Watt?

4. Oct 2, 2015

### srarahcha

So when I multiplied it, i got 40000J. When I divided instead, I got 6.4x10^21.

5. Oct 2, 2015

### srarahcha

so I just thought about how power (Watts) is Joules over a specific time interval.... so if the Power is 1.60 x 1013Watts over a 2.50 ns interval, and P=J/Δt, then if I rearrange the equation to 1.60 x 1013 W * 2.50 x 10-9 s = Joules, i would also get the joules/second (aka Watts with a time interval of 1 s) because Joules over 1 second is the same number....

6. Oct 2, 2015

### Staff: Mentor

One Watt is one Joule per second: 1 W = 1 J/s. So if you multiply by a given time (such as your 2.50 ns) you get the total energy in Joules produced over that time interval. The result has the unit Joules. It's then up to you to interpret what that value means in the context of what you're trying to accomplish. For example, you now know how much energy you can get from a single firing of the laser.